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I need a proof that -1 DOES NOT equal 1

by anis91
Tags: equal, falacy, math, negative one, proof, prove, sqrt
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anis91
#1
Nov22-13, 07:33 AM
P: 5
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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Office_Shredder
#2
Nov22-13, 07:47 AM
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We get this question a lot, there's actually a thread in the FAQ devoted to answering questions like it

http://www.physicsforums.com/showthread.php?t=637214

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.
anis91
#3
Nov22-13, 07:54 AM
P: 5
okay thanks alot! i appreciate it! ^^

quietrain
#4
Nov22-13, 08:03 AM
P: 651
I need a proof that -1 DOES NOT equal 1

Quote Quote by Office_Shredder View Post
We get this question a lot, there's actually a thread in the FAQ devoted to answering questions like it

http://www.physicsforums.com/showthread.php?t=637214

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!
Mark44
#5
Nov22-13, 09:14 AM
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P: 21,408
Quote Quote by quietrain View Post
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.
Quote Quote by quietrain View Post

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.
quietrain
#6
Nov22-13, 10:30 AM
P: 651
Quote Quote by Mark44 View Post
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 ?
Mark44
#7
Nov22-13, 07:40 PM
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P: 21,408
Quote Quote by quietrain View Post
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.
tade
#8
Nov22-13, 08:13 PM
tade's Avatar
P: 145
It'd be cooler if someone managed to "prove" that i=√1
quietrain
#9
Nov23-13, 05:43 AM
P: 651
Quote Quote by Mark44 View Post
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
Zeda
#10
Nov23-13, 08:38 PM
P: 10
Quote Quote by tade View Post
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]
jbriggs444
#11
Nov24-13, 06:46 AM
P: 999
Quote Quote by Zeda View Post
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.
hilbert2
#12
Nov24-13, 07:51 AM
P: 327
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/h1...al-numbers.pdf .

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).
1MileCrash
#13
Nov29-13, 02:05 PM
1MileCrash's Avatar
P: 1,305
Quote Quote by Zeda View Post
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.
Zeda
#14
Nov29-13, 02:18 PM
P: 10
Quote Quote by 1MileCrash View Post
√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.


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