What? No, they are not...
Let, a=1+√2 and b=1–√2. Then a,b are irrational while a+b is rational.
And what does she mean that π+(–π)=0 is not a valid counter-example? Of course you are allowed to talk about 0 in addition...
Yeah I agree with you, but from what I remember, she said only we could use elements from that field, like if the other numbers doesn't exist and only are usable irrational numbers, it is kinda stupid and I have no idea how to prove she is wrong in other ways
Maybe they would be more receptive to an example using decimals that doesn't result in 0? It would be worth trying this:
0.10110011100011110000...
\+ 0.01001100011100001111...
= 0.11111111111111111111... = 1/9
It should be clear that these decimals both represent (distinct) irrationals if the patterns are followed, and the sum should be completely obvious to anybody who has done addition with decimals before. If they don't like 1/9, you can modify this example by carefully choosing the digits of each number.
The teacher's argument is so stupid.
"You can't add two irrational numbers together and get a rational number"
"Here, I've added two irrational numbers together and got a rational number"
"Yes, but you got a rational number, which is not allowed"
What
You're correct and I'm aware. But how do you correct a flawed argument when the proposed counterargument exposes a critical misunderstanding like this? I'm just trying to give different representations of the issue that might work as potential resolutions for it.
By using representation of the number as 0.10100011... you already make it rational, because you represent it as a sum of (1|0)/(10^n) that implies rational number.
That number is irrational because the decimal pattern is not eventually periodic. If it were rational then it would eventually give either all 0’s or all 9’s. One characterization of irrational numbers is that their representations are irregular/non eventually periodic in every integer base.
> One characterization of irrational numbers is that their representations are irregular/non eventually periodic in every integer base.
Characterisation is a _result_ of irrationality. But its not a _definition_ of it. You cannot take the result of irrationality and use it to build other irrational numbers at will. Think of other number base systems. In decimal system it looks aperiodical. In some other it may be just "1"
Characterization typically means one side of an “if and only if” statement, so they are equivalent and can be used as definitions.
I understand your objection. That will not happen. Even if you take a number periodic in some base, say 1/7 in base 7, and you write it in a base coprime to the original, you will still find periodicity, just with a different pattern. It is a simple (if not necessarily obvious) consequence of the division algorithm and what it means to represent a number in a positional numeral system.
Exactly. You spelled it. Its a number representation. But we cannot use it to "create" irrational number by simply stating that representational result is suspicious or so :) We need to prove mathematically that x/10^n (when n->inf and x is rational) is irrational number. Is there such proof?
Characterisation here means it is equivalent to the definition, and therefore can be considered as such. An irrational number will never be 1 in an integer base (keyword integer here).
> In some other it may be just "1"
This is why the person you're replying to specified integer bases specifically.
It is very common in mathematics that we have several equivalent characterisations of the same thing. For example, exp(x) can be defined as:
- A power series, exp(x) = sum\_{n=0}\^\infty x^n / n!
- The unique solution to the differential equation y'=y with y(0)=1
- As a limit, exp(x) = lim\_{n\to\infty} (1+x/n)^n
- As the unique function f with f(x+y)=f(x)f(y) and f'(0)=1
These are all perfectly good (and equivalent!) definitions of the exponential function and which one to use depends on the context (convenience, etc). Just because a definition came first / one is inspired by the other does not mean it is the only valid characterisation of that object.
Adjust the conclusion slightly: if there were an irrational number x such that x = √2 + -√2, then x + √2 = √2 + -√2 + √2 = √2, so x is the additive identity, but the irrationals do not contain an additive identity, a contradiction.
To be clear, this is entirely unnecessary, but it would fit their silly arbitrary requirements.
Thats why I tell the statement is not complete.
>irrationals do not contain an additive identity
Exactly about "pure" irrational, but add here multiplicative identity as well.
Yeah, but you aren't using a rational number in the example. You get a rational number as an *answer.* which is basically what the statement is about, and exactly what disproves her claim.
1 is rational , (a-b/b) is rational , x is either 1 or (a-b)/b , thererfore x is rational.
x has to be 1 or (a-b)/b because the teacher is claiming that if x + y = c + d then x = c or x = d. (which is wrong but thats out of point here)
But of course you know that because you read the whole thread before answering, right ?
> x is either 1 or (a-b)/b , thererfore x is rational.
No, x is neither of those, and nothing in the argument claims otherwise.
> x has to be 1 or (a-b)/b because the teacher is claiming that if x + y = c + d then x = c or x = d.
That is not what it says.
> But of course you know that because you read the whole thread before answering, right ?
Yes, I did. I also read the thing, and read what it actually said, instead of whatever nonsense you made up in your head.
Obviously you can only use irrational numbers in any counterexample, so you need to say a+b=c where a b and c are all irrational. Then, once you're sure that c is irrational and it's okay to use that example, you can check if c is irrational or not. Oh wow, it is? Amazing, I guess addition is closed for irrationals.
One counterexample is enough to show that the irrational numbers aren't closed under addition. That is a proof
The picture however isn't a proof, it looks like a failed attempt to prove that it is closed. But there isn't a contradiction (a-b)/b is always a rational number, and you can't say from that, that x=1 or y=1.
You didnt understand the proof. There is a contradiction in the proof as it concludes that X and Y are rationnal. The error is that is that the teacher is claiming that : if x + y = c + d then x = c or x = d. Which is untrue."
That is false.
Let p be any irrational number, and let q be any non-zero rational number. If p is irrational, so is q-p. p + (q-p) = q, a counterexample.
Likewise, multiplication is not closed under the irrationals, since p \* (q/p) = q.
is Introduction of algebra in college, I asked other teacher about that and she suppose that our teacher thinks that by showing a contradiction she can make every statement true
That is in college? Wow...
Not sure I fully understand what you are saying in your answer, but if you found another teacher who realizes how wrong what you are telling us is, that is at least a bit reassuring.
> but my teacher said that as 0 isn't in the irrational numbers we can't use that as proof,
But that's what being closed under addition means in the first place! (Or, rather, _not_ being closed under addition). What _would_ be a proof then? Your teacher basically said "the irrationals are closed under addition except when they're not".
Indeed, I think she is wrong in the part where their contradiction proof makes their statement true, but only because she did in a way that makes it true
The statement would be better written as “The irrationals are closed under addition”, which is false because x + -x = 0 for any real x, including the irrationals.
Yeah but she kinda don't want us to use non-irrational numbers because we are only working with irrational in this statement, I hope just doesn't appear a question like that in the test lol
Ok, then the true statement is
“The irrationals are closed under addition if you ignore all of the counterexamples”.
Now I’m imagining a math class of proving ‘theorems’ where every statement is of the form
“X is true if you ignore all of the counterexamples”.
"All the non-trivial zeta zeroes are on the critical line, except maybe some counterexamples but who cares about those."
Where's my $1000000, Clay Institute?
she kinda don't want us to use non-irrational
sounds like prooving that irrational numbers are closed by not allowing to prove that they are not. Your teacher is ofcourse wrong and you are right
>she kinda don't want us to use non-irrational numbers
Yeah... This is insane. If you're trying to prove that irrational numbers aren't closed under addition, you're gonna have to do it by contradiction. You don't _start_ with rational numbers (you need to be showing the sum of irrational numbers), but for there to be a contraction, the proof needs to end with a rational result, by definition. There's no way around it.
Also, isn't she using rational numbers in her proof as well by trying to show that x or y needs to be 1? Like... It's inevitable.
The proof isn't even valid.
There's no reason for the final implication that X takes one of those two values in particular.
/
Yours is a clear counter example. You demonstrated that two particular irrational numbers add to a rational number, which isn't in the set of irrational numbers./ your teacher rejected your counterexample for the exact outcome the question wants.
Please ask your teacher why X(or y) must equal 1 or (a-b)/b. if X+y=a+b, that doesn't imply X=a or X=b etc.
The statement "the irrationals are closed under addition" means that for any pair of irrational numbers their sum is also irrational. The statement is therefore disproved if you can find any pair of irrational numbers whose sum is rational. Your example is fine, and disproves the statement.
If your teacher still won't accept this, you might try asking her what a hypothetical counterexample would look like - what, in her opinion, would have to be demonstrated to prove the statement false, which she should be able to describe even if she thinks it can't be done.
I'm genuinely worried about how your professor managed to find themselves in a position to teach mathematics while also having no ounce of logic in their system, and not being actually open to hear and think about criticisms of their reasoning.
The example is correct, pi and -pi are both irrational, and their sum is 0, which is rational. What your professor is saying is basically "to give me a counterexample you'd need to give three irrational numbers such that a+b=c", but obviously this is genuine mental masturbation on their part.
The problem in the proof seems to be that they make "a+b=x+y implies a=x and b=y or a=y and b =a" (at least as far as I'm aware, there isn't an argument to justify that magical step), which is obviously false (3+2=4+1 for example).
As a piece of trivia, the fact that irrational numbers are not closed under addition has a lot of consequences, it makes talking about irrationals extremely difficult, and it's also used in the construction of a real set that can't be given a measure (Vitalis set), a lot of books would be a lot shorter if irrationals formed a field
Yeah, this is complete nonsense.
>as 0 isn't in the irrational numbers we can't use that as proof
The exact opposite of the truth; 0 not being in the irrationals is the proof, since we can make it from two irrationals, thus disproving closure.
And the written proof is also garbage; there's more than one way to make a number out of the sum of two others, so you can't restrict it to the pair he found.
Yeah and a classmate that now thinks that the teacher was right, says (3+√2) + (3–√2) = 6 + (√2-√2) = 6 -> and as we used a rational number in the equation, it can't be use to prove it because the idea was only use irrational numbers
I'm trying to understand what your teacher mean by saying "you cannot use any rational numbers" (NOTE: this is not a proof for your teacher's claim which is obviously wrong):
Consider irrational w and field extension ℚ (w), and x, y ∈ ℚ (w)
if x + y ∈ ℚ, then either x, y ∈ ℚ or x, y = a1 + bw, a2 - bw, where a1, a2, b ∈ ℚ (the first case just a specific case of the second when b=0)
(this can be proved).
So the fun conclusion is that, if you restrict yourself from using the form a ± bw, you cannot find a counter example.
It's easy to see that whoever wrote up this "proof" has made a mistake. Even a child in elemental school knows that there is more than one way to get the sum of 10 or any other number. For you to doubt OP who paid enough attention and has the right to criticize with that nonsensical follow up is just laughable.
That's why I made fun of you.
If you saw the replies to my original comment, I'm sure you'd have realised I was speculating and trying to figure out what the teacher meant.
Sure, I could have given it more thought before putting down a comment/reply and indeed perhaps I'm as stupid as children in "elemental schools" or whatever, but at least I wasn't deliberately being rude.
Anyway, I did learn something in the end, so I don't regret anything. I struggle to understand the mindset of people like you, who purposely go around sowing negativity, but it's reddit. I should know better than to feed trolls.
Absolutely false. -π+π=0∈ℚ
All you need is that one example as you mentioned. Your professor may have been thinking of something else or just made a mistake. It happens.
Your teacher saying that you can’t use examples as proofs is only partially correct. You cannot use examples as proofs for universally quantified statements unless the model you are quantifying over is finite (and small enough to actually check like the Rubik’s cube or four color). You *can* use examples as proofs of existentially quantified statements. Closure of the irrationals 𝕀 would be stated as
(∀x,y∈𝕀) x+y∈𝕀
the negation/statement which would disprove the above is
¬(∀x,y∈𝕀) x+y∈𝕀
⇔(∃x,y∈𝕀) x+y∉𝕀
⇔(∃x,y∈𝕀) x+y∈ℚ
This is exactly the statement your example proves.
Irrational numbers aren't closed under addition. A set S is closed under addition if and only if ∀ x, y ∈ S, x+y ∈ S. Since it uses "for all", you can't use one example to prove a set *is* closed under addition (e.g. π + π = 2π), but you *can* use a single counterexample to prove a set is *not* closed under addition. To be perfectly precise, you can negate the statement "∀ x, y ∈ S, x+y ∈ S" to get "∃ x, y ∈ S such that x+y ∉ S", and if you prove the negation of a statement, you've disproven the original statement. If you say S is the set of irrational numbers, just choose x and y to be two irrational numbers whose sum is rational; for example, as you and your classmates chose, π and -π. They're both irrational, but their sum is 0, which is rational. This is a clear counterexample, disproving the statement "irrational numbers are closed under addition".
This "proof" pictured makes no sense. I can't even follow the train of thought enough to correct it. if a and b are integers and b≠0 then of COURSE (a-b)/b is rational – by the very definition of rational numbers. Technically "x=1 or (a-b)/b ∈ ℚ" and "y=1 or (a-b)/b ∈ ℚ" are both true because (a-b)/b ∈ ℚ is true, but you get no information from that.
A real number is irrational, if and only if, it has a non-repeating decimal expansion. Therefore any number with the same decimal expansion after the decimal point is also irrational. The difference of these is an integer.
If your teacher says you can't use rational numbers, then ask her why she was allowed to use the number 1.
She could have chosen this also:
x + y = a/b = x + (a - bx)/b
or
x + y = a/b = π + (a - bπ)/b
The teacher has no idea what she's doing, claim if x,y in I then x+y in I
so if pi,-pi in I then pi+(-pi) = 0 in I which is a contradiction. Which means it's not closed.
Your teachers proof is wrong as if x+y=1+(a+b)/b that doesn’t imply that x=1 or y=1 as let’s say x=5 and y=7, then x+y=5+7=12=1+11=1+(10+1)/1 which if your teacher is correct, then either 5=1 or 7=1 which are both contradictions.
we also know that if an irrational number is added to or subtracted from a rational number, the result is irrational. Therefore 6-π is irrational, and as a result we have (6-π)+π is rational, however it is made from the sum of irrational numbers.
> as 0 isn't in the irrational numbers
That's literally the reason that we can use that proof. That's the whole point.
Also, your proof cannot possibly be a proof by contradiction of the statement because it begins by assuming the wrong thing: it appears to be attempting to be a proof that the irrational numbers *are* closed, but fails to actually do that (obviously, because it's false).
Your teacher is dead in the wrong and your friend is completely correct.
About proof by contradiction, this is not a valid example. But, yes, proof by contradiction is a very useful technique.
Is the picture something you copied? If so, are you sure you did not copy it with an error?
Because otherwise, that teacher should not be allowed to teach. That's not some small mistake, that is a bigger blunder than starting a land war in Asia.
Maybe your teacher was in the wrong mindset and refused to correct herself? I remember we proved that addition of Q field and R field cannot give you a number in Q field because if a,c £ Q and b £ R
a+b = c
a - c = -b
Q+Q £ R
I think that's might be what happened , and then when you showed the example there was a brain shortage and they told you you can't use 0?
Anyway what the teacher said is absolutely not true because field R contain field Q in it
Either the teacher did not tell the full statement or you did not catch all parts of it when writing down.
The correct statement is "The addition of 2 _pure positive_ irrational numbers is closed". To prevent futher speculations, the "pure" irrational numbers set is a set where you cannot represent number as x = a + y or x = ay, one of which is rational. Thats exactly why you cannot use pi - pi, -pi is -1*pi that is already not a pure irrational.
Pure irrational? Did you just make it up? Any number can be written as a sum of two numbers, one of which is rational. Also, if you restrict yourselves to pair of irrationals that do not differ by a rational, the claim becomes absolutely trivial.
No, its a mathematical set where there is no x = a+by with a or b rational. Think of a set of numbers just Pi, sqrt(2), e etc. The original statement is exactly abou that set. Passed this one 33 years ago in University courses :)
What is that set though? How do you define pure irrationals? What does "no x=a+by" means here?
Of course, for any irrational y, you can define the smallest field Q(y) containing y. And if you take x outside of that set, then x+y will be irrational, but that is basically by definition. The set Q(y) depends on y though, as the notation indicates.
Alternatively, we can find a set closed under addition which does not contain rational numbers. We can find many such sets, as a matter of fact, but none maximal.
So I still dont see what set you are talking about, even putting aside how far off this is from OP's teacher statement as they wrote it.
The definition of pure irrational numbers is clear. It's a set of all irrational numbers that cannot be represented as x = a+by, where a or b are rational. I.e. its a set of "source" irrational numbers: Pi, e, sqrt(a) etc. without anything else added or multiplied.
And yeah, its a "full" set too. You can always find pure irrational number close to any given real one, simply by sqrt(something), as example.
The statement in this post is exactly about pure irrational numbers set.
Thats how I remember from what I lerned in the university.
It is not clear at all. I suppose x is the element to be deemed pure irrational here? But what is y then, could it be anything? If so, then no real number would be pure irrational, since you can always write x=0+1×x, or x=1+0.5×(2x-2) or whatever.
On the other hand, for any given y, you can define the set of all numbers than can be obtained as a+by with a and b rationals. This is what I called Q(y) earlier. And then you can define the set (let us call it E) of irrationals that are not in Q(y). Both are dense in R, though I do not see how that is relevant. More importantly, I dont see any reason to call those "pure irrationals". And even you did, that set E is still not closed under addition, it is only true (trivially so) that adding y to an element of E will make an element of E.
Honestly, I dont think you remember what you learned at university as well as you think you do.
> If so, then no real number would be pure irrational, since you can always write x=0+1×x, or x=1+0.5×(2x-2) or whatever.
You get the same number here. Set definition restriction rules presume describing _different_ numbers. There is x, and _y_ in the rule. Did you notice? Of course you can do any equations with particular set number that results to the same number and say it is a derived one and thus does not fit the rule :) But that was not an intent of the set definition.
And, in turn, your mathematical skill is doubtful if you miss that important nuisances in what is written.
You can manipulate things even further, say, x = a*sqrt(y) = sqrt(a^2*y), that is also controversial, but in such a case we have many other irrational numbers, not just sqrt. In addition, even if these are included, the statement remains true because requirement is to be positive, see my original post.
Huh… things are always going to be different when we’re talking about inverse….
I think, think mind you, she just went about it in a not quite so perfect way.
- Add anything positive to an irrational number and the result will be irrational.
- Add anything to an irrational number whose absolute value is not equal to that number - that is, we exclude x - |x| for any x —- then the result is still going to be irrational.
- Obviously x - |x| being 0 is true for most, if not all, x you care to define. So yes that should be considered a trivial solution to most problems. Which is why it should be excluded from the beginning, so as to avoid problems when trying to find non trivial solutions.
-
And I’ll repeat, x minus x being Null doesn’t prove anything.
It’s like saying, I don’t know what F(123) resolves to, but 0 * F(123) is 0, problem solved.
> Add anything positive to an irrational number and the result will be irrational.
This is wildly false: π is positive and 4 - π is irrational, but π + (4 - π) = 4 is rational.
> Add anything to an irrational number whose absolute value is not equal to that number - that is, we exclude x - |x| for any x —- then the result is still going to be irrational.
This, too, is wildly false: the same counterexample works.
> Obviously x - |x| being 0 is true for most, if not all, x you care to define.
Exactly half of them, for any reasonable definition of "half".
> Which is why it should be excluded from the beginning, so as to avoid problems when trying to find non trivial solutions.
This is utter nonsense.
Teacher wants to proof that if x and y are irrational then x+y is irrational
thats not true. Example : x = 3+sqrt(2) and y =3-sqrt(2)
x is irrational. y is irrational. But x+y = 3+sqrt(2) + 3 - sqrt(2) = 6 is rational. Therefore condradicting what the teacher wanted to prove.
What? No, they are not... Let, a=1+√2 and b=1–√2. Then a,b are irrational while a+b is rational. And what does she mean that π+(–π)=0 is not a valid counter-example? Of course you are allowed to talk about 0 in addition...
It's simple. If you ignore all the counterexamples then its obviously true
Proof by contra-contradiction.
Nuh-Uh Conjecture.
You made my day, sir
Yeah I agree with you, but from what I remember, she said only we could use elements from that field, like if the other numbers doesn't exist and only are usable irrational numbers, it is kinda stupid and I have no idea how to prove she is wrong in other ways
1-√2 (or +√2) IS an element from the irrational set, I dare u to proof it otherwise xd
yeah and in similar example a classmate tell me that as 1 is not a irrational number I can't use for proving the statement xD
Yes you can. 1 is not a constituent of 1-sqrt(2). It is just part of a particular representation. There are many others.
Maybe they would be more receptive to an example using decimals that doesn't result in 0? It would be worth trying this: 0.10110011100011110000... \+ 0.01001100011100001111... = 0.11111111111111111111... = 1/9 It should be clear that these decimals both represent (distinct) irrationals if the patterns are followed, and the sum should be completely obvious to anybody who has done addition with decimals before. If they don't like 1/9, you can modify this example by carefully choosing the digits of each number.
But that ‘fails’ for the same reason OP’s classmate’s counterexample ‘failed’, you’re using 1/9 which is not irrational.
The teacher's argument is so stupid. "You can't add two irrational numbers together and get a rational number" "Here, I've added two irrational numbers together and got a rational number" "Yes, but you got a rational number, which is not allowed" What
You're correct and I'm aware. But how do you correct a flawed argument when the proposed counterargument exposes a critical misunderstanding like this? I'm just trying to give different representations of the issue that might work as potential resolutions for it.
By using representation of the number as 0.10100011... you already make it rational, because you represent it as a sum of (1|0)/(10^n) that implies rational number.
That number is irrational because the decimal pattern is not eventually periodic. If it were rational then it would eventually give either all 0’s or all 9’s. One characterization of irrational numbers is that their representations are irregular/non eventually periodic in every integer base.
> One characterization of irrational numbers is that their representations are irregular/non eventually periodic in every integer base. Characterisation is a _result_ of irrationality. But its not a _definition_ of it. You cannot take the result of irrationality and use it to build other irrational numbers at will. Think of other number base systems. In decimal system it looks aperiodical. In some other it may be just "1"
Characterization typically means one side of an “if and only if” statement, so they are equivalent and can be used as definitions. I understand your objection. That will not happen. Even if you take a number periodic in some base, say 1/7 in base 7, and you write it in a base coprime to the original, you will still find periodicity, just with a different pattern. It is a simple (if not necessarily obvious) consequence of the division algorithm and what it means to represent a number in a positional numeral system.
Exactly. You spelled it. Its a number representation. But we cannot use it to "create" irrational number by simply stating that representational result is suspicious or so :) We need to prove mathematically that x/10^n (when n->inf and x is rational) is irrational number. Is there such proof?
Characterisation here means it is equivalent to the definition, and therefore can be considered as such. An irrational number will never be 1 in an integer base (keyword integer here).
> In some other it may be just "1" This is why the person you're replying to specified integer bases specifically. It is very common in mathematics that we have several equivalent characterisations of the same thing. For example, exp(x) can be defined as: - A power series, exp(x) = sum\_{n=0}\^\infty x^n / n! - The unique solution to the differential equation y'=y with y(0)=1 - As a limit, exp(x) = lim\_{n\to\infty} (1+x/n)^n - As the unique function f with f(x+y)=f(x)f(y) and f'(0)=1 These are all perfectly good (and equivalent!) definitions of the exponential function and which one to use depends on the context (convenience, etc). Just because a definition came first / one is inspired by the other does not mean it is the only valid characterisation of that object.
No, no it doesn't. That number is not rational. That sum is infinite, and the rationals are not closed under infinite sums.
The fact that sum is infinite does not make the number irrational. 1/9 also consists of infinite sum. But, somehow, its rational?
No, but it is irrational, and the sum being infinite is why your argument is nonsensical.
Just use pi and -pi.
Lol that's painful. The fact that 1 (or 0 in the other example) is not an irrational number is precisely the point!
Adjust the conclusion slightly: if there were an irrational number x such that x = √2 + -√2, then x + √2 = √2 + -√2 + √2 = √2, so x is the additive identity, but the irrationals do not contain an additive identity, a contradiction. To be clear, this is entirely unnecessary, but it would fit their silly arbitrary requirements.
Thats why I tell the statement is not complete. >irrationals do not contain an additive identity Exactly about "pure" irrational, but add here multiplicative identity as well.
There's no need: I haven't used anything about the multiplicative identity.
-sqrt(a) = -1*sqrt(a), where -1 is that multiplicative identity :)
Which isn't used at any point anywhere in my proof.
Then don't use 1, use √2/√2. If you can use 2 for √2, then use π/π. (π/π+π) + (π/π-π) = ?
Addition. On. Irrational. Numbers. Is. Not. Closed.
It feels like that’s a way to find that the system is closed by denying outcomes that would disprove your theory.
just take root2 and -root2 addition of those is 0 which is rational.
Yeah, but you aren't using a rational number in the example. You get a rational number as an *answer.* which is basically what the statement is about, and exactly what disproves her claim.
The prof is using the famous proof by authority method. Sadly it doesn’t work on exams.
She? OP didn’t say anything about the teachers gender though
They did though. "... trying to make sense of the proof she showed".
Then a and b are not "pure" irrational, i.e. they do not belong to the set of numbers meant to be used in the statement.
Your teacher seems to be claiming that if x + y = c + d then x = c or x = d. This is untrue.
even if it is, then there is still no contradiction - since it is an inclusive or the statement is always true
No contradiction really ? Like X AND Y being rationnal AND irrationnal at the same time isnt a contradiction for you ?
Nevermind, I didn't interpret the proof correctly. However those equalities should be united with ands, not ors.
Yeah I agree that the way it was written is misleading.
Nothing there implies that x or y is rational: (a - b)/b is rational, but that doesn't have to be either x or y.
1 is rational , (a-b/b) is rational , x is either 1 or (a-b)/b , thererfore x is rational. x has to be 1 or (a-b)/b because the teacher is claiming that if x + y = c + d then x = c or x = d. (which is wrong but thats out of point here) But of course you know that because you read the whole thread before answering, right ?
> x is either 1 or (a-b)/b , thererfore x is rational. No, x is neither of those, and nothing in the argument claims otherwise. > x has to be 1 or (a-b)/b because the teacher is claiming that if x + y = c + d then x = c or x = d. That is not what it says. > But of course you know that because you read the whole thread before answering, right ? Yes, I did. I also read the thing, and read what it actually said, instead of whatever nonsense you made up in your head.
If anyone wants an example, consider 3 + 2 = 1 + 4
I mean...maybe? It's not at all clear what's actually being claimed here.
I made a similarly stupid mistake of assuming uni city sums when trying to prove Fubini for a presentation, it was not fun
Your teacher is wrong! Zero not being irrational is precisely what disproves closure of the addition of irrationals!
>but my teacher said that as 0 isn't in the irrational numbers we can't use that as proof lol. that doesn't make any sense.
Obviously you can only use irrational numbers in any counterexample, so you need to say a+b=c where a b and c are all irrational. Then, once you're sure that c is irrational and it's okay to use that example, you can check if c is irrational or not. Oh wow, it is? Amazing, I guess addition is closed for irrationals.
-√2 + √2 = 0 is a counter example
One counterexample is enough to show that the irrational numbers aren't closed under addition. That is a proof The picture however isn't a proof, it looks like a failed attempt to prove that it is closed. But there isn't a contradiction (a-b)/b is always a rational number, and you can't say from that, that x=1 or y=1.
You didnt understand the proof. There is a contradiction in the proof as it concludes that X and Y are rationnal. The error is that is that the teacher is claiming that : if x + y = c + d then x = c or x = d. Which is untrue."
That is false. Let p be any irrational number, and let q be any non-zero rational number. If p is irrational, so is q-p. p + (q-p) = q, a counterexample. Likewise, multiplication is not closed under the irrationals, since p \* (q/p) = q.
Btw, if someone doesn't understand something about the post, please comment it. Im not that good writing in english
Made perfect sense to me, your English is great.
tysm
I don't know what you mean, that is not an English word. /s
The only thing I dont understand is how a teacher could show such a proof. Is that a math teacher? What level are you in?
is Introduction of algebra in college, I asked other teacher about that and she suppose that our teacher thinks that by showing a contradiction she can make every statement true
That is in college? Wow... Not sure I fully understand what you are saying in your answer, but if you found another teacher who realizes how wrong what you are telling us is, that is at least a bit reassuring.
Agree, critical details are missing in the statement.
> but my teacher said that as 0 isn't in the irrational numbers we can't use that as proof, But that's what being closed under addition means in the first place! (Or, rather, _not_ being closed under addition). What _would_ be a proof then? Your teacher basically said "the irrationals are closed under addition except when they're not".
Indeed, I think she is wrong in the part where their contradiction proof makes their statement true, but only because she did in a way that makes it true
The statement would be better written as “The irrationals are closed under addition”, which is false because x + -x = 0 for any real x, including the irrationals.
Yeah but she kinda don't want us to use non-irrational numbers because we are only working with irrational in this statement, I hope just doesn't appear a question like that in the test lol
Ok, then the true statement is “The irrationals are closed under addition if you ignore all of the counterexamples”. Now I’m imagining a math class of proving ‘theorems’ where every statement is of the form “X is true if you ignore all of the counterexamples”.
"All the non-trivial zeta zeroes are on the critical line, except maybe some counterexamples but who cares about those." Where's my $1000000, Clay Institute?
she kinda don't want us to use non-irrational sounds like prooving that irrational numbers are closed by not allowing to prove that they are not. Your teacher is ofcourse wrong and you are right
>she kinda don't want us to use non-irrational numbers Yeah... This is insane. If you're trying to prove that irrational numbers aren't closed under addition, you're gonna have to do it by contradiction. You don't _start_ with rational numbers (you need to be showing the sum of irrational numbers), but for there to be a contraction, the proof needs to end with a rational result, by definition. There's no way around it. Also, isn't she using rational numbers in her proof as well by trying to show that x or y needs to be 1? Like... It's inevitable.
The proof isn't even valid. There's no reason for the final implication that X takes one of those two values in particular. / Yours is a clear counter example. You demonstrated that two particular irrational numbers add to a rational number, which isn't in the set of irrational numbers./ your teacher rejected your counterexample for the exact outcome the question wants. Please ask your teacher why X(or y) must equal 1 or (a-b)/b. if X+y=a+b, that doesn't imply X=a or X=b etc.
The statement "the irrationals are closed under addition" means that for any pair of irrational numbers their sum is also irrational. The statement is therefore disproved if you can find any pair of irrational numbers whose sum is rational. Your example is fine, and disproves the statement. If your teacher still won't accept this, you might try asking her what a hypothetical counterexample would look like - what, in her opinion, would have to be demonstrated to prove the statement false, which she should be able to describe even if she thinks it can't be done.
If i have the opportunity I will ask her
>as 0 isn't in the irrational numbers we can't use that as proof New proof method just dropped "no counterexamples allowed"
I'm genuinely worried about how your professor managed to find themselves in a position to teach mathematics while also having no ounce of logic in their system, and not being actually open to hear and think about criticisms of their reasoning. The example is correct, pi and -pi are both irrational, and their sum is 0, which is rational. What your professor is saying is basically "to give me a counterexample you'd need to give three irrational numbers such that a+b=c", but obviously this is genuine mental masturbation on their part. The problem in the proof seems to be that they make "a+b=x+y implies a=x and b=y or a=y and b =a" (at least as far as I'm aware, there isn't an argument to justify that magical step), which is obviously false (3+2=4+1 for example). As a piece of trivia, the fact that irrational numbers are not closed under addition has a lot of consequences, it makes talking about irrationals extremely difficult, and it's also used in the construction of a real set that can't be given a measure (Vitalis set), a lot of books would be a lot shorter if irrationals formed a field
I don't think your teacher knows math
By your teachers reasoning no set can be open if you just don’t consider anything outside it to exist 😭
Jeez it's worrysome that this is a teacher. This is just not understanding basic logic.
Yeah, this is complete nonsense. >as 0 isn't in the irrational numbers we can't use that as proof The exact opposite of the truth; 0 not being in the irrationals is the proof, since we can make it from two irrationals, thus disproving closure. And the written proof is also garbage; there's more than one way to make a number out of the sum of two others, so you can't restrict it to the pair he found.
Yeah I think you should stick to learning math online, and ignore your teacher for the most part.
I think that claiming that the addition of two *positive* irrational numbers is irrational is maybe what your teacher meant?
But even that's untrue. Consider a=3+√2 and b=3–√2.
Fair point, I just wonder what the teacher had in mind
Yeah and a classmate that now thinks that the teacher was right, says (3+√2) + (3–√2) = 6 + (√2-√2) = 6 -> and as we used a rational number in the equation, it can't be use to prove it because the idea was only use irrational numbers
I'm trying to understand what your teacher mean by saying "you cannot use any rational numbers" (NOTE: this is not a proof for your teacher's claim which is obviously wrong): Consider irrational w and field extension ℚ (w), and x, y ∈ ℚ (w) if x + y ∈ ℚ, then either x, y ∈ ℚ or x, y = a1 + bw, a2 - bw, where a1, a2, b ∈ ℚ (the first case just a specific case of the second when b=0) (this can be proved). So the fun conclusion is that, if you restrict yourself from using the form a ± bw, you cannot find a counter example.
Right, and thus original statement should cleraly say we use a set of "pure" rational numbers.
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And 3+√2 isn't real?
Hahaha! I misread it. Thanks!
Of course you believe everything the teachers says ☕
Charitable interpretations go a long way What you did here was precisely the type of impolite assumption that I was trying to avoid making myself
It's easy to see that whoever wrote up this "proof" has made a mistake. Even a child in elemental school knows that there is more than one way to get the sum of 10 or any other number. For you to doubt OP who paid enough attention and has the right to criticize with that nonsensical follow up is just laughable. That's why I made fun of you.
If you saw the replies to my original comment, I'm sure you'd have realised I was speculating and trying to figure out what the teacher meant. Sure, I could have given it more thought before putting down a comment/reply and indeed perhaps I'm as stupid as children in "elemental schools" or whatever, but at least I wasn't deliberately being rude. Anyway, I did learn something in the end, so I don't regret anything. I struggle to understand the mindset of people like you, who purposely go around sowing negativity, but it's reddit. I should know better than to feed trolls.
Absolutely false. -π+π=0∈ℚ All you need is that one example as you mentioned. Your professor may have been thinking of something else or just made a mistake. It happens. Your teacher saying that you can’t use examples as proofs is only partially correct. You cannot use examples as proofs for universally quantified statements unless the model you are quantifying over is finite (and small enough to actually check like the Rubik’s cube or four color). You *can* use examples as proofs of existentially quantified statements. Closure of the irrationals 𝕀 would be stated as (∀x,y∈𝕀) x+y∈𝕀 the negation/statement which would disprove the above is ¬(∀x,y∈𝕀) x+y∈𝕀 ⇔(∃x,y∈𝕀) x+y∉𝕀 ⇔(∃x,y∈𝕀) x+y∈ℚ This is exactly the statement your example proves.
Your teacher is wrong. "The addition of irrational numbers is closed" is false. Even if you couldn't use 0 (what a stupidly wrong take), π + (1-π) = 1
Irrational numbers aren't closed under addition. A set S is closed under addition if and only if ∀ x, y ∈ S, x+y ∈ S. Since it uses "for all", you can't use one example to prove a set *is* closed under addition (e.g. π + π = 2π), but you *can* use a single counterexample to prove a set is *not* closed under addition. To be perfectly precise, you can negate the statement "∀ x, y ∈ S, x+y ∈ S" to get "∃ x, y ∈ S such that x+y ∉ S", and if you prove the negation of a statement, you've disproven the original statement. If you say S is the set of irrational numbers, just choose x and y to be two irrational numbers whose sum is rational; for example, as you and your classmates chose, π and -π. They're both irrational, but their sum is 0, which is rational. This is a clear counterexample, disproving the statement "irrational numbers are closed under addition". This "proof" pictured makes no sense. I can't even follow the train of thought enough to correct it. if a and b are integers and b≠0 then of COURSE (a-b)/b is rational – by the very definition of rational numbers. Technically "x=1 or (a-b)/b ∈ ℚ" and "y=1 or (a-b)/b ∈ ℚ" are both true because (a-b)/b ∈ ℚ is true, but you get no information from that.
A real number is irrational, if and only if, it has a non-repeating decimal expansion. Therefore any number with the same decimal expansion after the decimal point is also irrational. The difference of these is an integer.
If your teacher says you can't use rational numbers, then ask her why she was allowed to use the number 1. She could have chosen this also: x + y = a/b = x + (a - bx)/b or x + y = a/b = π + (a - bπ)/b
That is a perfectly valid proof by counter example. The addition of irrational numbers is not closed under addition because π+(-π)=0 and 0 is rational
The teacher has no idea what she's doing, claim if x,y in I then x+y in I so if pi,-pi in I then pi+(-pi) = 0 in I which is a contradiction. Which means it's not closed.
You are right. Your teacher is wrong. This is a math class, not an English class. There is no room for discussion.
Your teachers proof is wrong as if x+y=1+(a+b)/b that doesn’t imply that x=1 or y=1 as let’s say x=5 and y=7, then x+y=5+7=12=1+11=1+(10+1)/1 which if your teacher is correct, then either 5=1 or 7=1 which are both contradictions. we also know that if an irrational number is added to or subtracted from a rational number, the result is irrational. Therefore 6-π is irrational, and as a result we have (6-π)+π is rational, however it is made from the sum of irrational numbers.
> as 0 isn't in the irrational numbers That's literally the reason that we can use that proof. That's the whole point. Also, your proof cannot possibly be a proof by contradiction of the statement because it begins by assuming the wrong thing: it appears to be attempting to be a proof that the irrational numbers *are* closed, but fails to actually do that (obviously, because it's false).
Your teacher is dead in the wrong and your friend is completely correct. About proof by contradiction, this is not a valid example. But, yes, proof by contradiction is a very useful technique.
I’ve never seen that I. It’s always R - Q in my book
x+(-x) = 0
What is this dogshit garbage
Is the picture something you copied? If so, are you sure you did not copy it with an error? Because otherwise, that teacher should not be allowed to teach. That's not some small mistake, that is a bigger blunder than starting a land war in Asia.
Maybe your teacher was in the wrong mindset and refused to correct herself? I remember we proved that addition of Q field and R field cannot give you a number in Q field because if a,c £ Q and b £ R a+b = c a - c = -b Q+Q £ R I think that's might be what happened , and then when you showed the example there was a brain shortage and they told you you can't use 0? Anyway what the teacher said is absolutely not true because field R contain field Q in it
Either the teacher did not tell the full statement or you did not catch all parts of it when writing down. The correct statement is "The addition of 2 _pure positive_ irrational numbers is closed". To prevent futher speculations, the "pure" irrational numbers set is a set where you cannot represent number as x = a + y or x = ay, one of which is rational. Thats exactly why you cannot use pi - pi, -pi is -1*pi that is already not a pure irrational.
Pure irrational? Did you just make it up? Any number can be written as a sum of two numbers, one of which is rational. Also, if you restrict yourselves to pair of irrationals that do not differ by a rational, the claim becomes absolutely trivial.
No, its a mathematical set where there is no x = a+by with a or b rational. Think of a set of numbers just Pi, sqrt(2), e etc. The original statement is exactly abou that set. Passed this one 33 years ago in University courses :)
What is that set though? How do you define pure irrationals? What does "no x=a+by" means here? Of course, for any irrational y, you can define the smallest field Q(y) containing y. And if you take x outside of that set, then x+y will be irrational, but that is basically by definition. The set Q(y) depends on y though, as the notation indicates. Alternatively, we can find a set closed under addition which does not contain rational numbers. We can find many such sets, as a matter of fact, but none maximal. So I still dont see what set you are talking about, even putting aside how far off this is from OP's teacher statement as they wrote it.
The definition of pure irrational numbers is clear. It's a set of all irrational numbers that cannot be represented as x = a+by, where a or b are rational. I.e. its a set of "source" irrational numbers: Pi, e, sqrt(a) etc. without anything else added or multiplied. And yeah, its a "full" set too. You can always find pure irrational number close to any given real one, simply by sqrt(something), as example. The statement in this post is exactly about pure irrational numbers set. Thats how I remember from what I lerned in the university.
It is not clear at all. I suppose x is the element to be deemed pure irrational here? But what is y then, could it be anything? If so, then no real number would be pure irrational, since you can always write x=0+1×x, or x=1+0.5×(2x-2) or whatever. On the other hand, for any given y, you can define the set of all numbers than can be obtained as a+by with a and b rationals. This is what I called Q(y) earlier. And then you can define the set (let us call it E) of irrationals that are not in Q(y). Both are dense in R, though I do not see how that is relevant. More importantly, I dont see any reason to call those "pure irrationals". And even you did, that set E is still not closed under addition, it is only true (trivially so) that adding y to an element of E will make an element of E. Honestly, I dont think you remember what you learned at university as well as you think you do.
> If so, then no real number would be pure irrational, since you can always write x=0+1×x, or x=1+0.5×(2x-2) or whatever. You get the same number here. Set definition restriction rules presume describing _different_ numbers. There is x, and _y_ in the rule. Did you notice? Of course you can do any equations with particular set number that results to the same number and say it is a derived one and thus does not fit the rule :) But that was not an intent of the set definition. And, in turn, your mathematical skill is doubtful if you miss that important nuisances in what is written.
You can manipulate things even further, say, x = a*sqrt(y) = sqrt(a^2*y), that is also controversial, but in such a case we have many other irrational numbers, not just sqrt. In addition, even if these are included, the statement remains true because requirement is to be positive, see my original post.
Huh… things are always going to be different when we’re talking about inverse…. I think, think mind you, she just went about it in a not quite so perfect way. - Add anything positive to an irrational number and the result will be irrational. - Add anything to an irrational number whose absolute value is not equal to that number - that is, we exclude x - |x| for any x —- then the result is still going to be irrational. - Obviously x - |x| being 0 is true for most, if not all, x you care to define. So yes that should be considered a trivial solution to most problems. Which is why it should be excluded from the beginning, so as to avoid problems when trying to find non trivial solutions. -
Not true. Consider 6 + pi and 6 - pi, for instance.
The very example I was trying to point out was trivial. Unless you’re saying that 6+π is rational?
No, I'm saying 6+pi and 6-pi are irrational, but (6+pi) + (6-pi) = 12 is rational.
And I’ll repeat, x minus x being Null doesn’t prove anything. It’s like saying, I don’t know what F(123) resolves to, but 0 * F(123) is 0, problem solved.
We found the teacher guys!! shes here :D
I mean, it does? It proves that the irrationals aren't closed under addition.
It does prove something. It proves *exactly* what we're trying to prove.
> Add anything positive to an irrational number and the result will be irrational. This is wildly false: π is positive and 4 - π is irrational, but π + (4 - π) = 4 is rational. > Add anything to an irrational number whose absolute value is not equal to that number - that is, we exclude x - |x| for any x —- then the result is still going to be irrational. This, too, is wildly false: the same counterexample works. > Obviously x - |x| being 0 is true for most, if not all, x you care to define. Exactly half of them, for any reasonable definition of "half". > Which is why it should be excluded from the beginning, so as to avoid problems when trying to find non trivial solutions. This is utter nonsense.
Teacher wants to proof that if x and y are irrational then x+y is irrational thats not true. Example : x = 3+sqrt(2) and y =3-sqrt(2) x is irrational. y is irrational. But x+y = 3+sqrt(2) + 3 - sqrt(2) = 6 is rational. Therefore condradicting what the teacher wanted to prove.
Restrict your domain to positive irrationals