I need a proof that -1 DOES NOT equal 1

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In summary, the conversation discusses a common question about an equation that seems to prove that -1 equals 1. However, it is explained that the equation is invalid because it violates the rules for taking exponents of negative numbers. The conversation also delves into the issue of even and odd functions, and how they can produce multiple outputs for a single input. Finally, it is stated that a rigorous proof for why -1 does not equal 1 can be found by using the axioms of real numbers.
  • #1
anis91
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hey everybody, once i saw a thread here (didn't want to revive it) about an equation that proves that 1=-1, it was proved wrong ofc, but at the end, someone posted this:

" -1=(-1)^1
=(-1)^2*1/2
=[(-1)^2]^1/2
=(1)^1/2
=√1
=1 "
yet no one replied to it, can someone show me which is the "trippy" step here? the one that misuses an algebra rule? (e.g. a rule that can only be applied to positive numbers etc..."

thank you.
 
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  • #2
We get this question a lot, there's actually a thread in the FAQ devoted to answering questions like it

https://www.physicsforums.com/showthread.php?t=637214 [Broken]

The main point is that whne you write
[tex] \left( -1 \right)^{2/2} = \left( (-1)^2 \right)^{1/2} [/tex]
you have performed an operation which is not actually valid. Taking exponents of negative numbers is tricky and you have to be more careful than when you are working with exponents of positive numbers. In general
[tex] x^{ab} =\left( x^{a} \right)^{b} [/tex]
is something that can only be applied when x is a positive number.
 
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  • #3
okay thanks alot! i appreciate it! ^^
 
  • #4
Office_Shredder said:
We get this question a lot, there's actually a thread in the FAQ devoted to answering questions like it

https://www.physicsforums.com/showthread.php?t=637214 [Broken]

The main point is that whne you write
[tex] \left( -1 \right)^{2/2} = \left( (-1)^2 \right)^{1/2} [/tex]
you have performed an operation which is not actually valid. Taking exponents of negative numbers is tricky and you have to be more careful than when you are working with exponents of positive numbers. In general
[tex] x^{ab} =\left( x^{a} \right)^{b} [/tex]
is something that can only be applied when x is a positive number.

hello office_shredder,

may i know why the indices rule is invalid for negative numbers? i tried for example, (-2^6) and split them up to [-2^(2*3)] = 4^3 and i still yielded 64.

where does taking exponents of negative numbers breakdown?

thanks!
 
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  • #5
quietrain said:
hello office_shredder,

may i know why the indices rule is invalid for negative numbers? i tried for example, (-2^6) and split them up to [-2^(2*3)] = 4^3 and i still yielded 64.
No it doesn't. The parentheses you have in (-2^6) don't do anything and might as well not be there. (-2^6) is exactly the same as -2^6 which is the same as -(2^6) or -64.

If you want to raise -2 to the 6th power, you have to write it as (-2)^6.
quietrain said:
where does taking exponents of negative numbers breakdown?
It breaks down when the exponent is fractional and represents an even root (i.e., square root, fourth root, and so on). There is no problem when the exponent is an integer unless you happen to be taking 0 to a negative power.
 
  • #6
Mark44 said:
No it doesn't. The parentheses you have in (-2^6) don't do anything and might as well not be there. (-2^6) is exactly the same as -2^6 which is the same as -(2^6) or -64.

If you want to raise -2 to the 6th power, you have to write it as (-2)^6.
It breaks down when the exponent is fractional and represents an even root (i.e., square root, fourth root, and so on). There is no problem when the exponent is an integer unless you happen to be taking 0 to a negative power.

yes that was sloppy of me :D

anyway, is the even root fractional exponent the only case whereby this rule breaks down ?
 
  • #7
quietrain said:
anyway, is the even root fractional exponent the only case whereby this rule breaks down ?
Yes, since odd roots (cube root, fifth root, and so on) can have negative arguments. For example, ##\sqrt[3]{-27} = -3## and ##\sqrt[5]{-32} = -2##.

If you have an expression such as (-27)2/3, you can write it either as [(-27)2]1/3 or as [(-27)1/3]2, both of which are equal to 9.

The first expression simplifies to (729)1/3 = 9, and the second expression simplifies to (-3)2, which is also 9.
 
  • #8
It'd be cooler if someone managed to "prove" that i=√1
 
  • #9
Mark44 said:
Yes, since odd roots (cube root, fifth root, and so on) can have negative arguments. For example, ##\sqrt[3]{-27} = -3## and ##\sqrt[5]{-32} = -2##.

If you have an expression such as (-27)2/3, you can write it either as [(-27)2]1/3 or as [(-27)1/3]2, both of which are equal to 9.

The first expression simplifies to (729)1/3 = 9, and the second expression simplifies to (-3)2, which is also 9.

that was very insightful, thank you
 
  • #10
tade said:
It'd be cooler if someone managed to "prove" that i=√1
I won't try that, but I thought of this yesterday (breaking the same rule as above):

Start with:
[itex]\sqrt{x}[/itex]
Now to factor out a -1:
[itex]=i\sqrt{-x}[/itex]
And to factor out another -1:
[itex]=i*i\sqrt{x}[/itex]
[itex]=-\sqrt{x}[/itex]

[itex]\Rightarrow \sqrt{x}=-\sqrt{x}[/itex]

:P The issue is that even functions are not 1-1, meaning they can map multiple inputs to the same output. Naturally, the inverse function would have to map backwards, but it would have to be split off to multiple values. That is why, for example, [itex]\sqrt{9}=\{3,-3\}[/itex]
 
  • #11
Zeda said:
That is why, for example, [itex]\sqrt{9}=\{3,-3\}[/itex]

It is a standard notational convention that [itex]\sqrt{x}[/itex] where x is a non-negative real number always refers to the positive root.
 
  • #12
i need a proof that -1 DOES NOT equal 1

If you want a rigorous proof for this kind of statements, you need to use the axioms of real numbers: http://math.berkeley.edu/~talaska/h1b/axioms-real-numbers.pdf [Broken] .

First you add ##1## to both sides of the equation ##1=-1## and get ##1+1=0##. Next you use the order axioms to show that ##1+1>1>0##, which is a contradiction and proves that ##1## can't equal ##-1## (for real numbers ##a## and ##b##, the inequalities ##a>b## and ##a=b## can't both be true).
 
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  • #13
Zeda said:
I won't try that, but I thought of this yesterday (breaking the same rule as above):

Start with:
[itex]\sqrt{x}[/itex]
Now to factor out a -1:
[itex]=i\sqrt{-x}[/itex]
And to factor out another -1:
[itex]=i*i\sqrt{x}[/itex]
[itex]=-\sqrt{x}[/itex]

[itex]\Rightarrow \sqrt{x}=-\sqrt{x}[/itex]

:P The issue is that even functions are not 1-1, meaning they can map multiple inputs to the same output. Naturally, the inverse function would have to map backwards, but it would have to be split off to multiple values. That is why, for example, [itex]\sqrt{9}=\{3,-3\}[/itex]

√9 is 3, √9 is not -3, and √x is a mapping from one real to precisely oneother real; a function.
 
  • #14
1MileCrash said:
√9 is 3, √9 is not -3, and √x is a mapping from one real to precisely oneother real; a function.

I know, I should have been more clear by being more confusing :P I was using √ to represent the function that, given the output of f(x)=x2, would return x. When I was working on a little project dealing with sine and cosine, I would often have a function squared on one side, where the otherside, after taking the square root, was indeed negative and the positive square root would cause the fully reduced form to fail.
 

What is the concept of equality in mathematics?

The concept of equality in mathematics is that two quantities or expressions are considered equal if they have the same value. This means that they can be substituted for each other in any mathematical equation or expression without changing the overall result.

Why is it important to prove that -1 does not equal 1?

Proving that -1 does not equal 1 is important because it ensures the accuracy and validity of mathematical statements and equations. If we assume that -1 equals 1, it can lead to incorrect solutions and false conclusions.

What is the proof that -1 does not equal 1?

The proof that -1 does not equal 1 is based on the properties of multiplication and the definition of equality. We know that when two numbers are multiplied, the result is positive unless one of the numbers is negative. Therefore, if -1 were equal to 1, -1*(-1) would equal 1*(-1), and this would result in both sides of the equation being equal to -1. However, we know that -1*(-1) equals 1, which contradicts the assumption that -1 equals 1. Thus, we can conclude that -1 does not equal 1.

Can -1 ever equal 1 in any mathematical context?

No, -1 can never equal 1 in any mathematical context. As mentioned before, the properties of multiplication and the definition of equality do not allow for this to happen. In any equation or expression, -1 and 1 will always have different values and cannot be considered equal.

Are there any real-life examples that demonstrate that -1 does not equal 1?

Yes, there are several real-life examples that demonstrate that -1 does not equal 1. For instance, if we have a debt of $1 and we owe someone $1, the total amount we owe is -1+1=0, not -1*1=1. Another example is the concept of negative temperature, where -1 and 1 represent two different states of matter (solid and gas) and cannot be considered equal.

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