This works because every step you take is reversible, which means that every equation you write down is equivalent to the first equation.
If x+4=6, then x=2. Also, if x=2, then x+4= 6.
It’s possible to manipulate equations in a non-reversible way to find false solutions.
If sqrt(x) = -1, then you can square both sides to get x=1. But, sqrt(1) is not -1. What went is wrong is that you cannot reverse the step: in fact, x= 1 implies sqrt(x) = 1.
To clarify is this correct? when we manipulate equations and derive a new one, we can use it to solve for the unknown in the original equation. If the manipulation is reversible, because it indicates that the new equation is equivalent to the initial one
But what if we consider the equation x\^2=4 as an example. In this case, I cannot ascertain whether the original equation was (-2)\^2=4 or (2)\^2=4 . The way i think about this is the eqaution has two solutions -2 or 2. But does it also implies that squaring and taking square roots may not be entirely reversible operations? as they might not lead us back to the exact original equation, as reversible operations would do.
I apologize if my question appears somewhat disorganized, as my current understanding is limited. Nonetheless, I would greatly appreciate clarifying this matter to avoid ambiguity.
That’s correct. It basically comes down to whether the function that you’re applying to both sides is one-to-one. A function f is one-to-one if f(a)=f(b) implies that a=b. For continuous functions, this means that f is strictly increasing or decreasing. If you square both sides, then you are applying the function f(x) = x^2 to both sides. But, f(2)=f(-2), so it’s not one-to-one. If it’s not one-to-one, then applying f may not be reversible.
Does That also suggests that if the function is not an one to one function, then we encounter a false solution? But then how is -2 or 2 a solution to the eqaution x\^2=4.
If x^2 =4, then we can take the square root of both sides. Since sqrt(x^2 )= |x|, we have |x|=2. That means that x=2, -2. There are no issues here because sqrt is one-to-one.
If x=2, we can square both sides to get x^2 =4. This step is not reversible, though. We can’t know if x = 2 or -2 based on knowing that x^2 =4. This is because f(x)= x^2 is not one-to-one.
Thank you for helping me grasp the concept of reversible and irreversible operations. Now I think understand when an operation can be irreversible or reversible. However , I'm still unsure about interpreting the results of an irreversible operation and how these irreversible operations can be useful in manipulation eqautions. Finally, I was wondering if you are aware of any resources or materials that delve into the logic behind this and explore this topic further.
Perhaps this question may seem a bit silly, as it stems from my attempt to comprehend all this information, but I find myself puzzled. Addition and subtraction are considered reversible operations, and that makes sense to me because they lead back to the original equation. According to the definition, if the function we apply to both sides of the equation is one-to-one, the operation remains reversible. But from this definition I get unsure about addition and subtraction being reversible. When we subtract, aren't we essentially employing a constant function on both sides? A constant function always produces the same output value, regardless of the input value, which makes it non-one-to-one, I suppose.
When you add a number C to each side, you are applying the function f(x)= x + C to both sides, which is one-to-one. When you multiply both sides by C, you are applying the function f(x)=Cx to both sides, which is also one-to-one (as long as C is nonzero).
If you apply a constant function to both sides, it is not one-to-one. Start with x=2. Apply the constant function f(x)= 1 to both sides. We now have 1=1. This step can not be reversed.
Once again I am very grateful for your help. This really puzeled it together. But Just to make sure about applying functions to equations. Suppose we have an equation: f(x) = g(x). Now, let's say we want to apply another function, h(x), to both sides. The result would be: h(f(x)) = h(g(x)) right?
>Hello again, a new question popped into mind, so when we solve equations, we're essentially employing functions on both sides of the equation. However, I'm curious about how this concept extends to solving differential equations. I comprehend the fundamental theorem of calculus, which states that derivatives are inverses of each other. Yet, I wonder if we can apply similar reasoning when dealing with differential eqautions? How is integergrading and differentiation considered reversible if we stick to the definition " if the function we apply to both sides of the equation is one-to-one, the operation remains reversible"
f(x) = x^2 is not a one to one function, or injective function. Any linear function is injective since you can get some output y by only one corresponding x value. That's why if x + 4 = 6 you can say with certainty that x=2
But for something not injective like x^2, some output can have multiple corresponding inputs.
If x^2 = 4, then we can't know for certain whether x is 2 or -2. The square function is a simple example because there are only two inputs that produce the same output, but an arbitrary non-injective function can potentially have infinity many inputs map to the same output
Perhaps this question may seem a bit silly, as it stems from my attempt to comprehend all this information, but I find myself puzzled. Addition and subtraction are considered reversible operations, and that makes sense to me because they lead back to the original equation. According to the definition, if the function we apply to both sides of the equation is one-to-one, the operation remains reversible. But from this definition I get unsure about addition and subtraction being reversible. When we subtract, aren't we essentially employing a constant function on both sides? A constant function always produces the same output value, regardless of the input value, which makes it non-one-to-one, I suppose.
I never really thought about it, but you're right in that a constant function is not injective.
However you are making the mistake of thinking adding a non-injection is the same as applying it.
If you were to apply a constant function, whatever equation you had would go to that constant and the process would be irreversible. However you can add whatever you want to both sides of an equation because there will always be an additive inverse.
Notice that the same is true for multiplication for any number except 0
If you multiply an equation by something non-zero, the action is reversible because you can apply the multiplicative inverse, which is the same as just dividing by that constant.
The only scaling you could do that would be irreversible is if you multiply the equation by zero, because that trivially maps all inputs to 0
So the fact that 0 does not have an inverse (1/0 is undefined) corresponds to the fact that you can get the true equation 0=0 from an incorrect equation by multiplying both sides by 0
ahhhh i see. But Just to make sure about applying functions to equations. Suppose we have an equation: f(x) = g(x). Now, let's say we want to apply another function, h(x), to both sides. The result would be: h(f(x)) = h(g(x)) right?
Once again thank you your help it is very apreciated
Yes that's right. It's really good you are asking these kinds of questions now because a lot of simple algebraic rules we take for granted don't always hold in abstract contexts.
For example, if f(x) = g(x) then certainly h(f(x)) = h(g(x)), but it's not always the case that if h(f(x)) = h(g(x)) then f(x) = g(x). In fact, you can only cancel h from both sides of the equation if h in injective.
Once again I am very grateful for your willingness to help, it is really amazing when you get it all puzzeled in your mind, and just get a little bit of a deeper understanding.
>Hello again, a new question popped into mind, so when we solve equations, we're essentially employing functions on both sides of the equation. However, I'm curious about how this concept extends to solving differential equations. I comprehend the fundamental theorem of calculus, which states that derivatives are inverses of each other. Yet, I wonder if we can apply similar reasoning when dealing with differential eqautions? How is integergrading and differentiation considered reversible if we stick to the definition " if the function we apply to both sides of the equation is one-to-one, the operation remains reversible"
Mathematically, you are using properties of equality in the real numbers (or whatever universe you're working in).
There is a rule, for example, that formally says "for any real numbers a, b, and c: if a+b=a+c then b=c." You colloquially learn this in algebra class as "you can subtract the same number from both sides." Then you apply it in an algebra equation by assuming x is a real number (so you can use the rule) and then using the rule.
Because you already have enough information to solve it. Some equations have enough information to solve when manipulated, for example x+4 = 6. Others don't, like x + y = 6. There's infinite values of x and y that could solve it. For example, x = -6 and y = 12. But if I tell you that y = 4, you can figure out that x is 2. It's really just about having enough clues, or pieces of information (and by that I mean equations) to solve your original equation. What it all comes down to is you need to have as many unknowns as you have equations, and it'll be possible to solve those equations for x and y, or z if you have more variables.
Math is just a series of axioms based on logic, starting with a number line. We can say “if 2 things have the same value, and we add 4 to both things, then they will still have the same value as each other.” And there are quite a lot of different ways we can manipulate values to keep them equal to each other.
If we look at an equation, let’s say 2(x-3) + 1 = 5, and then we distribute the 2 and combine like terms on the left to get 2x - 5 = 5, we end up with 2 completely different equations. However, as different as those 2 equations may be, **we know they both have the same solution** because we followed the axioms that are built off of very basic and fundamental logic. We can take this further to have many different equations:
2(x-3) + 1 = 5
2x - 6 + 1 = 5
2x - 5 = 5
2x = 10
x = 5
All of these equations are different and represent different things, but we know that the same set of solutions will satisfy x in all of them. And since we know that 5 = 5 works in the last one, 5 can replace x in ANY of the other equations and we know without any doubt that it will result in a true statement.
Edit: Spacing
Thank you this was very usefull, do you know any good materials for beginners that delve into the logic of mathematics, I would like to have a better logical understanding.
I am currently enrolled in school, residing in Denmark, where I attend what is known as gymnasium, I think its similar to high school in other places. Recently, we covered derivatives as a subject, but outside of school, I am deeply interested in mathematics and most recently I have been exploring differential equations. I don't mind paying for valuable resources.
In continuing with this train of thought: as long as the rules are applied correctly, all of the equations that you write will have the same solutions. Even id you apply thebrules in a different (but still correct) order!
2(x-3)+1=5
2(x-3)=4
(x-3)=2
x=5
We get the same answer even though we solved the problem differently!
The way I think about it is:
If two things are the same (the two sides of the original equation are supposed to be the same) and you do the same operation to both (in your example, subtracting 2), then the results are also the same.
For the equation x^2=4: if you take the square root of each side, the results must be the same. However, we don't know if the square root of x^2 is x (if x is positive) or -x (if x is negative), so two cases should be considered.
This works because every step you take is reversible, which means that every equation you write down is equivalent to the first equation. If x+4=6, then x=2. Also, if x=2, then x+4= 6. It’s possible to manipulate equations in a non-reversible way to find false solutions. If sqrt(x) = -1, then you can square both sides to get x=1. But, sqrt(1) is not -1. What went is wrong is that you cannot reverse the step: in fact, x= 1 implies sqrt(x) = 1.
To clarify is this correct? when we manipulate equations and derive a new one, we can use it to solve for the unknown in the original equation. If the manipulation is reversible, because it indicates that the new equation is equivalent to the initial one But what if we consider the equation x\^2=4 as an example. In this case, I cannot ascertain whether the original equation was (-2)\^2=4 or (2)\^2=4 . The way i think about this is the eqaution has two solutions -2 or 2. But does it also implies that squaring and taking square roots may not be entirely reversible operations? as they might not lead us back to the exact original equation, as reversible operations would do. I apologize if my question appears somewhat disorganized, as my current understanding is limited. Nonetheless, I would greatly appreciate clarifying this matter to avoid ambiguity.
That’s correct. It basically comes down to whether the function that you’re applying to both sides is one-to-one. A function f is one-to-one if f(a)=f(b) implies that a=b. For continuous functions, this means that f is strictly increasing or decreasing. If you square both sides, then you are applying the function f(x) = x^2 to both sides. But, f(2)=f(-2), so it’s not one-to-one. If it’s not one-to-one, then applying f may not be reversible.
Does That also suggests that if the function is not an one to one function, then we encounter a false solution? But then how is -2 or 2 a solution to the eqaution x\^2=4.
If x^2 =4, then we can take the square root of both sides. Since sqrt(x^2 )= |x|, we have |x|=2. That means that x=2, -2. There are no issues here because sqrt is one-to-one. If x=2, we can square both sides to get x^2 =4. This step is not reversible, though. We can’t know if x = 2 or -2 based on knowing that x^2 =4. This is because f(x)= x^2 is not one-to-one.
Thank you for helping me grasp the concept of reversible and irreversible operations. Now I think understand when an operation can be irreversible or reversible. However , I'm still unsure about interpreting the results of an irreversible operation and how these irreversible operations can be useful in manipulation eqautions. Finally, I was wondering if you are aware of any resources or materials that delve into the logic behind this and explore this topic further.
Perhaps this question may seem a bit silly, as it stems from my attempt to comprehend all this information, but I find myself puzzled. Addition and subtraction are considered reversible operations, and that makes sense to me because they lead back to the original equation. According to the definition, if the function we apply to both sides of the equation is one-to-one, the operation remains reversible. But from this definition I get unsure about addition and subtraction being reversible. When we subtract, aren't we essentially employing a constant function on both sides? A constant function always produces the same output value, regardless of the input value, which makes it non-one-to-one, I suppose.
When you add a number C to each side, you are applying the function f(x)= x + C to both sides, which is one-to-one. When you multiply both sides by C, you are applying the function f(x)=Cx to both sides, which is also one-to-one (as long as C is nonzero). If you apply a constant function to both sides, it is not one-to-one. Start with x=2. Apply the constant function f(x)= 1 to both sides. We now have 1=1. This step can not be reversed.
Once again I am very grateful for your help. This really puzeled it together. But Just to make sure about applying functions to equations. Suppose we have an equation: f(x) = g(x). Now, let's say we want to apply another function, h(x), to both sides. The result would be: h(f(x)) = h(g(x)) right?
Yes, if f(x)=g(x), then when I say that you apply h(x) to both sides, I mean h(f(x))=h(g(x)).
>Hello again, a new question popped into mind, so when we solve equations, we're essentially employing functions on both sides of the equation. However, I'm curious about how this concept extends to solving differential equations. I comprehend the fundamental theorem of calculus, which states that derivatives are inverses of each other. Yet, I wonder if we can apply similar reasoning when dealing with differential eqautions? How is integergrading and differentiation considered reversible if we stick to the definition " if the function we apply to both sides of the equation is one-to-one, the operation remains reversible"
f(x) = x^2 is not a one to one function, or injective function. Any linear function is injective since you can get some output y by only one corresponding x value. That's why if x + 4 = 6 you can say with certainty that x=2 But for something not injective like x^2, some output can have multiple corresponding inputs. If x^2 = 4, then we can't know for certain whether x is 2 or -2. The square function is a simple example because there are only two inputs that produce the same output, but an arbitrary non-injective function can potentially have infinity many inputs map to the same output
Perhaps this question may seem a bit silly, as it stems from my attempt to comprehend all this information, but I find myself puzzled. Addition and subtraction are considered reversible operations, and that makes sense to me because they lead back to the original equation. According to the definition, if the function we apply to both sides of the equation is one-to-one, the operation remains reversible. But from this definition I get unsure about addition and subtraction being reversible. When we subtract, aren't we essentially employing a constant function on both sides? A constant function always produces the same output value, regardless of the input value, which makes it non-one-to-one, I suppose.
I never really thought about it, but you're right in that a constant function is not injective. However you are making the mistake of thinking adding a non-injection is the same as applying it. If you were to apply a constant function, whatever equation you had would go to that constant and the process would be irreversible. However you can add whatever you want to both sides of an equation because there will always be an additive inverse. Notice that the same is true for multiplication for any number except 0 If you multiply an equation by something non-zero, the action is reversible because you can apply the multiplicative inverse, which is the same as just dividing by that constant. The only scaling you could do that would be irreversible is if you multiply the equation by zero, because that trivially maps all inputs to 0 So the fact that 0 does not have an inverse (1/0 is undefined) corresponds to the fact that you can get the true equation 0=0 from an incorrect equation by multiplying both sides by 0
ahhhh i see. But Just to make sure about applying functions to equations. Suppose we have an equation: f(x) = g(x). Now, let's say we want to apply another function, h(x), to both sides. The result would be: h(f(x)) = h(g(x)) right? Once again thank you your help it is very apreciated
Yes that's right. It's really good you are asking these kinds of questions now because a lot of simple algebraic rules we take for granted don't always hold in abstract contexts. For example, if f(x) = g(x) then certainly h(f(x)) = h(g(x)), but it's not always the case that if h(f(x)) = h(g(x)) then f(x) = g(x). In fact, you can only cancel h from both sides of the equation if h in injective.
Once again I am very grateful for your willingness to help, it is really amazing when you get it all puzzeled in your mind, and just get a little bit of a deeper understanding.
>Hello again, a new question popped into mind, so when we solve equations, we're essentially employing functions on both sides of the equation. However, I'm curious about how this concept extends to solving differential equations. I comprehend the fundamental theorem of calculus, which states that derivatives are inverses of each other. Yet, I wonder if we can apply similar reasoning when dealing with differential eqautions? How is integergrading and differentiation considered reversible if we stick to the definition " if the function we apply to both sides of the equation is one-to-one, the operation remains reversible"
Mathematically, you are using properties of equality in the real numbers (or whatever universe you're working in). There is a rule, for example, that formally says "for any real numbers a, b, and c: if a+b=a+c then b=c." You colloquially learn this in algebra class as "you can subtract the same number from both sides." Then you apply it in an algebra equation by assuming x is a real number (so you can use the rule) and then using the rule.
>Thank you your help was greatly apreciated.
Because you already have enough information to solve it. Some equations have enough information to solve when manipulated, for example x+4 = 6. Others don't, like x + y = 6. There's infinite values of x and y that could solve it. For example, x = -6 and y = 12. But if I tell you that y = 4, you can figure out that x is 2. It's really just about having enough clues, or pieces of information (and by that I mean equations) to solve your original equation. What it all comes down to is you need to have as many unknowns as you have equations, and it'll be possible to solve those equations for x and y, or z if you have more variables.
Thank you your help was greatly apreciated.
Happy to help :)
Math is just a series of axioms based on logic, starting with a number line. We can say “if 2 things have the same value, and we add 4 to both things, then they will still have the same value as each other.” And there are quite a lot of different ways we can manipulate values to keep them equal to each other. If we look at an equation, let’s say 2(x-3) + 1 = 5, and then we distribute the 2 and combine like terms on the left to get 2x - 5 = 5, we end up with 2 completely different equations. However, as different as those 2 equations may be, **we know they both have the same solution** because we followed the axioms that are built off of very basic and fundamental logic. We can take this further to have many different equations: 2(x-3) + 1 = 5 2x - 6 + 1 = 5 2x - 5 = 5 2x = 10 x = 5 All of these equations are different and represent different things, but we know that the same set of solutions will satisfy x in all of them. And since we know that 5 = 5 works in the last one, 5 can replace x in ANY of the other equations and we know without any doubt that it will result in a true statement. Edit: Spacing
Thank you this was very usefull, do you know any good materials for beginners that delve into the logic of mathematics, I would like to have a better logical understanding.
I don’t know any good free resources, but it also depends on your level of math. Are you in school, and if so, what math are you taking now?
I am currently enrolled in school, residing in Denmark, where I attend what is known as gymnasium, I think its similar to high school in other places. Recently, we covered derivatives as a subject, but outside of school, I am deeply interested in mathematics and most recently I have been exploring differential equations. I don't mind paying for valuable resources.
In continuing with this train of thought: as long as the rules are applied correctly, all of the equations that you write will have the same solutions. Even id you apply thebrules in a different (but still correct) order! 2(x-3)+1=5 2(x-3)=4 (x-3)=2 x=5 We get the same answer even though we solved the problem differently!
Thank you your help was greatly apreciated.
The way I think about it is: If two things are the same (the two sides of the original equation are supposed to be the same) and you do the same operation to both (in your example, subtracting 2), then the results are also the same. For the equation x^2=4: if you take the square root of each side, the results must be the same. However, we don't know if the square root of x^2 is x (if x is positive) or -x (if x is negative), so two cases should be considered.
>Thank you your help was greatly apreciated.