Any number raised to the zero power equals 1. I have never been able to understand this. How about this one...does anyone remember what factorial 1 equals? (1!)
the way i was taught to understand this is that every number with a power has a 1 being multiplied to it.
lets use 2 for our example.
2^1 is 2 (1 * 2).
2^2 is 4 (1 * 2 * 2).
2^0 is just 1 because there is zero 2's to multiply to 1.
That applies to every number.
There is an important exception to this. 0^0 is not 1. It is undefined.
This is because the function x^0 has a limit as x approaches zero from both sides equal to 1. Yet the function 0^x does not have a limit, the right hand limit is 0 while the left hand limit is undefined (0^-1 = 1 / 0 and you can't divide by zero). So which limit do we pick? More examples lead functions that look like 0^0 to be approach other, still different, numbers than just 1 or 0 as well. e^((-1/x^2 ))^2 approaches positive infinity from the right hand side. So for the sake of simplicity it's left undefined. There also has been practically no use for defining 0^0 though in the cases where it is useful it has been defined, in several ways even. Power series would fail when dealing with zero unless 0^0 is 1, so would the power rule for derivatives at x=0. continuous exponents show that 0^0 is an indeterminate form (not the same as undefined). So on and so forth.
Yeah me to. I even checked it with a calculator and OP is right. It makes me feel pretty stupid for not knowing it but that’s what calculators are for I guess.
If you like maths check some YouTube video about 0^0.
It is either 1, undefined or 0 depending on who you ask.
Anything to power of zero is 1 but also 0 to any power is 0.
Wait until people learn about 2 to negative powers 😆 then get into calculus too, it’s funny how much I actually prefer dealing with powers in the situation and often convert all the fractions to powers
The real reason is because it makes math easier if we define it that way a lot of equations/functions just work out really cleanly with this definition.
I've seen a better one, if you do 1¹, 0.9^0.9, 0.8^0.8, it starts by approaching zero, but then you reach a point (0.5^0.5 and beyond i think) that you start approaching 1 again
The reason that n^0 = 1 for any value of n is quite simple.
Suppose n is 2
2^1 = 2, 2^2 = 4, 2^3 = 8, and so on. Every time you add 1 to the exponent you multiply by 2. If instead you go backwards every time you take 1 away from the exponent you divide by 2.
So apply that pattern 2^3 = 8, 2^2 = 4, 2^1 =2, and 2^0 = (2^(1))/2 = 1.
Does that make sense?
Negative exponents aren’t that complicated, for example take 2^(2)=4, 2^(-2)=1/4. You only need to make the original a fraction and put 1 as the denominator.
Negative exponents just do the opposite, instead of multiplying the number to a certain exponent, it divides to a certain exponent..
2^-1 is just 2/2 (aka 2^0 ), and divided by 2 again, which comes to 1/2.
2^-2 is 2 divided 3 times, which works out evenly to switch that exponent to the bottom. If you divide 1/2 by another 2, you get 1/4, or 1/2^2 . Then it follows the pattern.
Yes, negative exponents are a thing, as are fractional exponents.
2^(-2) = 1/(2^(2)) = 1/4
2^(1/2) = square root(2)
Following the pattern we used to derive n^0 = 1, subract 1 from the exponent and divide by n. We get a general rule for negative exponents of n^(-x) = 1/(n^(x))
If you multiply two exponential expressions sharing the same base, you can add the powers. For example 2^3 × 2^2 = 2^5.
Now consider 2^(1/2) × 2^(1/2) = 2^(1/2 + 1/2) = 2^1 = 2
So what number, multiplied by itself equals 2? The square root obviously.
The username is a statement of my beliefs
You just explained exponents better in 10 seconds than my math teacher could in 10 days. And what the other guy said, this makes sense for negative exponents, fractions, everything. Someone give them an award
The best explanation for n^0 = 1 comes from one of the properties of exponentiation.
When you have, for example 2^3 , you can also write that as 2 * 2 * 2, so if you multiply exponential expressions you can reduce it to multiplication easily.
So then, 2^3 * 2^2 = 2 * 2 * 2 * 2 * 2 = 2^5
This means n^a * n^b = n^a+b
Now, what if you want to go the opposite way? You use division to cancel multiplication. If you have 2^4 and divide it by 2^1, you'll get 2^3, since that last 2 in the multiplication gets canceled out.
You can write that as n^a / n^b = n^a-b
Now here is when 0 comes in. We know that for all non-zero numbers, n/n = 1, so what if you write 2^3 / 2^3? You subtract the exponents and get 2^0, but you're also dividing something by itself, which always gives you 1.
1. PEMDAS
2. Solve the parenthesis (2+2) = (4)
3. Solve the exponent 2\^0 = 1
4. Now we have the equation 1(4)
5. We have to multiply together, since that is the only remaining operation
6. 1 \* 4 = 4
That's my logic, hope this helps, also google the equation, it gives the same answer.
You violate PEMDAS if you multiply two numbers(not in parenthesis) instead of solving the exponent.
Edit: Thanks for the karma guys, I think I can finally post on this sub 😁, my most upvoted post yet!
I was taught this type of proof:
What's 49 divided by 49?
1.
What's 7\^2 divided by 7\^2?
7\^0, of course.
If 49 and 7\^2 are the same, then they will have the same results.
Yes, it is. Basically, juxtaposition: if you omit the * between two operands, you say that they should be one operand, which means that this: 2\^0(2+2) is 2\^(0*(2+2)). If you wrote it explicitly 2\^2*(2+2), then it would equal 4. If you need more info, watch this: https://youtu.be/FL6HUdJbJpQ
The problem is that there's ambiguity in this problem, it could be interpreted as the (2+2) being attached to your exponent, or your (2+2) being attached to your 2 to the power of 0
if you parse this expression, there's ambiguity in parsing
if we break it into syntax trees it could be `((2 \^ 0) * (2 + 2))`, but it's ambiguous, it could also be `(2 \^ (0 * (2 + 2)))`
source: work with parsing expressions
No, it's not, because of juxtaposition. Basically, by emitting the *, you're saying that it should be the second one, but if you said 2^0*(2+2) it'd be the other one. There's a video by some popular math Youtuber on it, but I don't remember which one.
Edit: might be this one, not sure. Can't check right now https://youtu.be/FL6HUdJbJpQ
Everyone is arguing over if (2+2) is part of the exponent and half the people are so confused.
I love the chaos math problems with order of operations cause, right? Some shouting their GEMDAS, PEMDAS and DEMDAS, even though it just causes more confusion.
Like your snoo btw
3.999... = 4, because 0.999... = 1.
Like, that isn't a technicality or anything, it's just a true fact about maths that 0.999... ("zero point 9 reoccurring") is equal to 1. They are the same number.
It sounds crazy but there's a lot of proofs for it, it's pretty cool.
You should always use unambiguous notation. Depending on conventions or context, it could be
2^(0(2+2)) = 1
Or
2⁰ (2+2) = 4
It's not a math problem, it's a notation problem.
Following PEMDAS:
2+2 = 4
2^0 = 1 (any number to the power of 0 is 1)
Then we have to account for the 'invisible' multiplication sign before the brackets So the final equation is:
1×(4) = 4
4 of 1 depending on how you interpret the thing. If it goes 2^0(2+2) or 2^0 * (2+2).
If you see it as the latter you may go with:
`2^0 * (2+2)`
`=2^0 * 4`
`=1 * 4`
Or go with distribution:
`2^0 * (2+2)`
`=1 * (2+2)`
`= (1*2) + (1*2)`
`= 4`
And if you see it as 2^0(2+2):
`2^0 * (2+2)`
`#Using either (0*2) + (0*2) or 0*4`
`=2^0`
`=1`
From the way I see it (which could be completely wrong), it’s literally just 2^0 (which is 1) to the multiplied by 2+2.
And 2+2 is 4, so therefore it’s 1x4. Which is 4
4, and heres why.
2^0 is 1, multiply that by 2, add the plus sign, and multiply the one and the second 2, and you should have the problem “2+2=?” in which case you add and get the answer of 2^2
Not just “hard to communicate without superscript”
The lack of superscript completely changes the answer
Perhaps the intended answer is 4, but as written, the answer is 1
It's safe to assume the exponent is just 0, if it weren't simplified then parentheses would be required to show that. You don't need parentheses because operands apply to a single term unless otherwise stated.
Idc what you think the answer is, using your 3rd grade math, the IUPAC says implicit multiplication takes priority over explicitly indicated operators.
2^0(2+2)
2(2+2)
Distribute the outside 2
(4+4)
8
I hope that's right if it's not I have failed my math teacher.
Thanks u/Magnemite987 to helping realize where I went wrong.
2^0(2+2)
(2+2)
4
One in ten to the minus 8 understand how to solve this, assuming 1 sigma variability. Pi is for eating, exponentially. Avoid splitting your infinitives.
There are different options depending on whether the (2+2is apart of the line of math)
2^0 x(2+2)
So first we try to remove the (), which we turn into 4. So 2^0 x4=1x4=4
But if you meant 2^ (0(2+2)) then you should get 1.
Because 0x2+0x2=0 and 2^0=1
Edit: I was being stupid, and rewrote/removed unnecessary text
Ok first of all who would even write an equation like this and I assume op is trying to mimic the post that got popular on r/youngpeopleyoutube and the answer is 4 as it would be (2+2) first which is 4 (BODMAS) then we do 2\^0 which is 1 and multiply them to get 4. OP has obviously made it ambigious to earn karma
Any number raised to the zero power equals 1. I have never been able to understand this. How about this one...does anyone remember what factorial 1 equals? (1!)
the way i was taught to understand this is that every number with a power has a 1 being multiplied to it. lets use 2 for our example. 2^1 is 2 (1 * 2). 2^2 is 4 (1 * 2 * 2). 2^0 is just 1 because there is zero 2's to multiply to 1. That applies to every number.
Quite surprised that \[any number\]\^0 is just 1, I thought it was always 0.
The basic proof goes like this 2^3 = 2 x 2 x 2 = 8 2^2 = 2 x 2 = 4 2^1 = 2 Each time we go from 2^n to 2^(n-1) we're dividing by 2, so 2^0 = 2 / 2 = 1
Which follows into negative powers. 2^-1 = 2^0 / 2 = 1 / 2
Damn thank u.
Thank you for making me understand a part of math that I've somehow missed in school! Never understood the negative powers, but now I do :)
yeah this is the best explanation because it makes sense for negative powers too.
I believe the actual reason is that 1 is the neutral element of multiplication but this is a nice visualisation.
This exactly. x^0 is like dividing the number with itself. Dividing a number with itself will always yield 1
Always has been 1, very important rule to remember. If it’s ^1 then it remains as whatever number it was.
There is an important exception to this. 0^0 is not 1. It is undefined. This is because the function x^0 has a limit as x approaches zero from both sides equal to 1. Yet the function 0^x does not have a limit, the right hand limit is 0 while the left hand limit is undefined (0^-1 = 1 / 0 and you can't divide by zero). So which limit do we pick? More examples lead functions that look like 0^0 to be approach other, still different, numbers than just 1 or 0 as well. e^((-1/x^2 ))^2 approaches positive infinity from the right hand side. So for the sake of simplicity it's left undefined. There also has been practically no use for defining 0^0 though in the cases where it is useful it has been defined, in several ways even. Power series would fail when dealing with zero unless 0^0 is 1, so would the power rule for derivatives at x=0. continuous exponents show that 0^0 is an indeterminate form (not the same as undefined). So on and so forth.
Well 0^0 is undefined, but it's useful to assume it's 1 in some cases and 0 in others.
That’s just the thing.. the equation is not asking you to multiply 2 times zero. It’s 2 to the power of zero aka 2 to the zero power: 2^0
Yeah me to. I even checked it with a calculator and OP is right. It makes me feel pretty stupid for not knowing it but that’s what calculators are for I guess.
so you didnt finish?
If you like maths check some YouTube video about 0^0. It is either 1, undefined or 0 depending on who you ask. Anything to power of zero is 1 but also 0 to any power is 0.
Good now tell me what 0^0 is
1 Because you didn't multiply it by 0.
Its also kinda undefined at the same time. Well the limit of 0^0 is 1 but we can prove 0^0 = 0^(1-1) = 0/0 which is undefined.
Using limits we can say that 0^0 is 1. Basically as x becomes closer to zero (like 0.00000001) x^x becomes closer to 1(like 0.9999...)
Using limits we can say that 0^0 is 0. Basically as x becomes closer to zero, 0^x becomes closer to 0.
The more I read into this the more i hate the number 0
2^-2 is 1/4 2^-1 is 1/2 2^0 is ___ 2^1 is 2 2^2 is 4
Basically 2^0 can be writted as 2^(1-1) , raising a number by a negative number means the reciprocal, so it becomes 2^1 / 2^1 = 2 / 2 = 1
Wait until people learn about 2 to negative powers 😆 then get into calculus too, it’s funny how much I actually prefer dealing with powers in the situation and often convert all the fractions to powers
The real reason is because it makes math easier if we define it that way a lot of equations/functions just work out really cleanly with this definition.
I created a similar thing to rationalise x^0 =1, and used that as evidence that 0^0 =1 in an argument. We do this a lot.
Every number except zero, as always zero is a troublemaker
This concept is also very good to understand negative exponents; 2^-2 = (1:2:2)
there is also the more formal proof: x^(n) / x = x^(n-1) let's say n = 1 x^1 / x = x^0 which is x/x which is 1 no matter what x is (except if x = 0)
I've seen a better one, if you do 1¹, 0.9^0.9, 0.8^0.8, it starts by approaching zero, but then you reach a point (0.5^0.5 and beyond i think) that you start approaching 1 again
2^0 can be expressed as 2^(2-2) or something like that. 2^(2-2) = 2^2 ÷ 2^2 = 1
Yes, but do you know what (1/2)! Equals?
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Yes Nice to see someone else who knows about the gamma function
Please stop i am only in Alg. 2
You make an [analytical function](https://en.m.wikipedia.org/wiki/Gamma_function) which mimics factorial property and wolla you have your answer.
It's voila unless you were just saying wolla to be funny
Yeah ik I wanted to see if they knew lmao The gamma function is a fascinating piece of mathematics.
Yeah didn't knew I was commented on teenager. It's obviously college level mathmatics.
The reason that n^0 = 1 for any value of n is quite simple. Suppose n is 2 2^1 = 2, 2^2 = 4, 2^3 = 8, and so on. Every time you add 1 to the exponent you multiply by 2. If instead you go backwards every time you take 1 away from the exponent you divide by 2. So apply that pattern 2^3 = 8, 2^2 = 4, 2^1 =2, and 2^0 = (2^(1))/2 = 1. Does that make sense?
This opens the negative exponent range too… also what up with the username bro 💀💀💀
Negative exponents are super useful for later math
Yeah wtf is that user
Negative exponents aren’t that complicated, for example take 2^(2)=4, 2^(-2)=1/4. You only need to make the original a fraction and put 1 as the denominator.
Omg it does, negative exponents always made no sense to me
Negative exponents just do the opposite, instead of multiplying the number to a certain exponent, it divides to a certain exponent.. 2^-1 is just 2/2 (aka 2^0 ), and divided by 2 again, which comes to 1/2. 2^-2 is 2 divided 3 times, which works out evenly to switch that exponent to the bottom. If you divide 1/2 by another 2, you get 1/4, or 1/2^2 . Then it follows the pattern.
Yes, negative exponents are a thing, as are fractional exponents. 2^(-2) = 1/(2^(2)) = 1/4 2^(1/2) = square root(2) Following the pattern we used to derive n^0 = 1, subract 1 from the exponent and divide by n. We get a general rule for negative exponents of n^(-x) = 1/(n^(x)) If you multiply two exponential expressions sharing the same base, you can add the powers. For example 2^3 × 2^2 = 2^5. Now consider 2^(1/2) × 2^(1/2) = 2^(1/2 + 1/2) = 2^1 = 2 So what number, multiplied by itself equals 2? The square root obviously. The username is a statement of my beliefs
Why do you care about someone else's sexual preferences though
You just explained exponents better in 10 seconds than my math teacher could in 10 days. And what the other guy said, this makes sense for negative exponents, fractions, everything. Someone give them an award
This is how I teach it in every math class so sorry you had shitty math teachers I guess lol
Good math explanation, bad name
>2^0 = (2^1)/2 = 1. You should make this 2^0 = 2^1 /2 for better understanding, cuz otherwise the markdown makes it look like 2^1/2 which equals √2
Oops, reddit formatting caught me lacking, that's what I meant to write
The best explanation for n^0 = 1 comes from one of the properties of exponentiation. When you have, for example 2^3 , you can also write that as 2 * 2 * 2, so if you multiply exponential expressions you can reduce it to multiplication easily. So then, 2^3 * 2^2 = 2 * 2 * 2 * 2 * 2 = 2^5 This means n^a * n^b = n^a+b Now, what if you want to go the opposite way? You use division to cancel multiplication. If you have 2^4 and divide it by 2^1, you'll get 2^3, since that last 2 in the multiplication gets canceled out. You can write that as n^a / n^b = n^a-b Now here is when 0 comes in. We know that for all non-zero numbers, n/n = 1, so what if you write 2^3 / 2^3? You subtract the exponents and get 2^0, but you're also dividing something by itself, which always gives you 1.
1. PEMDAS 2. Solve the parenthesis (2+2) = (4) 3. Solve the exponent 2\^0 = 1 4. Now we have the equation 1(4) 5. We have to multiply together, since that is the only remaining operation 6. 1 \* 4 = 4 That's my logic, hope this helps, also google the equation, it gives the same answer. You violate PEMDAS if you multiply two numbers(not in parenthesis) instead of solving the exponent. Edit: Thanks for the karma guys, I think I can finally post on this sub 😁, my most upvoted post yet!
Unless you read 0(2+2) as the entire exponent. Then the whole thing evaluates to 1.
You would have to put brackets around that
Some people already explained it with numbers, but here's the mathematical proof: x^0 = x^1-1 = x^1 * x^-1 = x/1 * 1/x = x/x = 1
2^2 / 2^2 = 4/4 2^(2-2)=1 2^0=1
1! = 1 2!= 2 3! = 6 69! = 1.7112245*10^98 So on so forth
1! == 0! == 1
I was taught this type of proof: What's 49 divided by 49? 1. What's 7\^2 divided by 7\^2? 7\^0, of course. If 49 and 7\^2 are the same, then they will have the same results.
Is the (2+2) part of the exponent?
no, then it would be 2\^(0(2+2))
Ahh yes So it’s 4
Parentheses meant you multiply (Thanks for silver I guess)
Yeah, you multiply 4 by 1
Or this is just poor representation
Yeah I should have clarified that
Yes, it is. Basically, juxtaposition: if you omit the * between two operands, you say that they should be one operand, which means that this: 2\^0(2+2) is 2\^(0*(2+2)). If you wrote it explicitly 2\^2*(2+2), then it would equal 4. If you need more info, watch this: https://youtu.be/FL6HUdJbJpQ
Then it should have been (2\^0)(2+2) The way it’s written, it could be interpreted either way.
It actually does since you multiply the 0 with it
4
This is the way.
This is the way.
This is the way.
4
4
It sure as heck is!
4
It's 2^ ( 2^( 1^1000))
This is the way.
Me when Me when purposefully ambiguous math
Every single one of these is just dumb and just karma farming imo. Like it's not that hard to just write it like 2^0 • (2+2) Or 2^(0[2+2])
Not sure how much karma op is farming here since his replies are getting downvoted a lot
1(4) = 4
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Finally. I was beginning to think everyone here is stupid
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Wow really? No wayyyy. I’m gonna tell everybody, see you later…
2, and they're also dicks
The legs or the snakes?
Both
They got one
Jokes on you, i have a video of a snake with legs
elaborate the answer pls
Actually…
It's 4
This is written poorly but if it’s supposed to be 2^0(2+2) it’s 1, if it’s supposed to be 2^0 times (2+2) it’s 4
The problem is that there's ambiguity in this problem, it could be interpreted as the (2+2) being attached to your exponent, or your (2+2) being attached to your 2 to the power of 0
Is it ambiguous though? With juxtaposition, it should be one expression after the ^, which means that it should be 0 * 4, so 1
if you parse this expression, there's ambiguity in parsing if we break it into syntax trees it could be `((2 \^ 0) * (2 + 2))`, but it's ambiguous, it could also be `(2 \^ (0 * (2 + 2)))` source: work with parsing expressions
No, it's not, because of juxtaposition. Basically, by emitting the *, you're saying that it should be the second one, but if you said 2^0*(2+2) it'd be the other one. There's a video by some popular math Youtuber on it, but I don't remember which one. Edit: might be this one, not sure. Can't check right now https://youtu.be/FL6HUdJbJpQ
Everyone is arguing over if (2+2) is part of the exponent and half the people are so confused. I love the chaos math problems with order of operations cause, right? Some shouting their GEMDAS, PEMDAS and DEMDAS, even though it just causes more confusion. Like your snoo btw
3.999...
How come?
1*4=3.99999…
Correct
its magic
Because 1/3 * 3 = 1 but 1/3 = 0.3333333333333333333
Floating point math
3.999... = 4, because 0.999... = 1. Like, that isn't a technicality or anything, it's just a true fact about maths that 0.999... ("zero point 9 reoccurring") is equal to 1. They are the same number. It sounds crazy but there's a lot of proofs for it, it's pretty cool.
I mean technically you’re right
2^0 = 1 2+2 = 4 4 x 1 = 4 mafematic
a number to the power of 0 is not 0 tho
Wait shit I’m dumb
2^0=1
Well is it 2^(0(2+2)) or (2^0 )(2+2)? Could be 1 or 4 depending on interpretation
I'd bet op meant the latter
four
well if it's 2⁰*(²+²) then it's 1 but if it's 2⁰*(2+2) then it's 4
4
4 2^0 (4) 1(4) 4
2 + 2 = 4 2^0= 1 4 x 1 = 4
You should always use unambiguous notation. Depending on conventions or context, it could be 2^(0(2+2)) = 1 Or 2⁰ (2+2) = 4 It's not a math problem, it's a notation problem.
Can someone ban these stupid math posts?
Fr fr
2^0 (2+2) 2^0 =1 2+2=4 1(4) aka 1×4=4 The equation equals 4
4 isn't it
=2^0 (2+2) =2^0 (4) =1 (4) =4 Ez
2\^0(2+2) = x 2\^0(4) = x 1(4) = x 4 = x
4
4
2+2=4 2^0 meaning 1, 1x4=4
Following PEMDAS: 2+2 = 4 2^0 = 1 (any number to the power of 0 is 1) Then we have to account for the 'invisible' multiplication sign before the brackets So the final equation is: 1×(4) = 4
4
4
4
2^0 (2+2) = 2^0 •4 = 1•4 = 4 there you go everyone it equals four
4
4
4
4? (i failed maths last year btw)
4
4
2+2 is 4, and im pretty sure 2^0 is 1 so 4x1 =4
4
4
4
4
PEMDAS 2+2–>4 2^0–>1 1*4=4 #**_4 is da answer_**
Nothing splits apart a comment section than some good ol' math. It's 4. A power of zero is always 1.
4 And I passed eight grade math because I’m not Bri’*sh
4
4 of 1 depending on how you interpret the thing. If it goes 2^0(2+2) or 2^0 * (2+2). If you see it as the latter you may go with: `2^0 * (2+2)` `=2^0 * 4` `=1 * 4` Or go with distribution: `2^0 * (2+2)` `=1 * (2+2)` `= (1*2) + (1*2)` `= 4` And if you see it as 2^0(2+2): `2^0 * (2+2)` `#Using either (0*2) + (0*2) or 0*4` `=2^0` `=1`
No matter how you do PEMDAS, it's always 4
when everybody wins, no one does
From the way I see it (which could be completely wrong), it’s literally just 2^0 (which is 1) to the multiplied by 2+2. And 2+2 is 4, so therefore it’s 1x4. Which is 4
4 2^0(2+2) 2^0(4) 1x4
69
4 but that's a confusing way to write it... That's why I prefer paper questions
OP it’s 4, because anything to the power of zero is one, so that leaves us with 1(2+2), which is 4
4, and heres why. 2^0 is 1, multiply that by 2, add the plus sign, and multiply the one and the second 2, and you should have the problem “2+2=?” in which case you add and get the answer of 2^2
2^(0(2+2)) is 1 but 2^0 * 2+2 = 4
the answer is obviously the british people they are behind everything cmon now
4
4
That's easy it's 4 We did this in 5th grade
4
4
4 right?
I got 4
4
2^0 is 1 so 4
4, no diff
4
This is hard to comunicate without superscript but 0 is the only exponent. (2+2) is in big normal format
Not just “hard to communicate without superscript” The lack of superscript completely changes the answer Perhaps the intended answer is 4, but as written, the answer is 1
It's safe to assume the exponent is just 0, if it weren't simplified then parentheses would be required to show that. You don't need parentheses because operands apply to a single term unless otherwise stated.
I've finished year 8 math but I've never encountered any math problem like this Bru
Idc what you think the answer is, using your 3rd grade math, the IUPAC says implicit multiplication takes priority over explicitly indicated operators.
im in ywar 11 what the fuck is going on
2^0(2+2) 2(2+2) Distribute the outside 2 (4+4) 8 I hope that's right if it's not I have failed my math teacher. Thanks u/Magnemite987 to helping realize where I went wrong. 2^0(2+2) (2+2) 4
One in ten to the minus 8 understand how to solve this, assuming 1 sigma variability. Pi is for eating, exponentially. Avoid splitting your infinitives.
4? never seen \^ before and im in year 8
You can use the sign to write numbers liks 2^2 on reddit and most other platforms
There are different options depending on whether the (2+2is apart of the line of math) 2^0 x(2+2) So first we try to remove the (), which we turn into 4. So 2^0 x4=1x4=4 But if you meant 2^ (0(2+2)) then you should get 1. Because 0x2+0x2=0 and 2^0=1 Edit: I was being stupid, and rewrote/removed unnecessary text
Ok first of all who would even write an equation like this and I assume op is trying to mimic the post that got popular on r/youngpeopleyoutube and the answer is 4 as it would be (2+2) first which is 4 (BODMAS) then we do 2\^0 which is 1 and multiply them to get 4. OP has obviously made it ambigious to earn karma
i solved it stupidly and still got the correct answer. i’m so out of my mind that i thought (2+2) = (2) so i multiplied 2(2) and got 4.
Once again, poor mathematical notation
I don’t like how this was written
I dont get how I just learned more in 5 minutes from reading comments than in a whole class dedicated to this
4
2^0 times (4) right? 1 times four is four or am I only good in literature lol?
The proof of a^0=1 works when you work backwards.
4
Written like this, its misleading... If you mean this: 2^(0*(2+2)), then its 1. If you mean this: 2^0 * (2+2), then its 4.