Why does x^n equal y^n when n is odd?

[Strants]

Homework Statement

From Spivak, Chapter 1, problem 7
Prove that if xⁿ = yⁿ and n is odd, then x=y. Hint: First explain why it suffices to consider only the case x, y >0; then show that x < y and x > y are both impossible.

Homework Equations

x > y means x-y \in P
x = y means x-y = 0
\mbox{Either}\: a \in P, -a \in P, \mbox{or}\: a=0
x^{n} - y^{n} = (x-y)(x^{n-1} + x^{n-2}y + \ldots + xy^{n-n2} + y^{n-1})
If n is odd, n = 2k+1

The Attempt at a Solution

I feel like this proof is right, at least in spirit. However, I'd like to check to see if it is rigorous enough.

If x > 0, then x^{2k+1}&gt;0. However, this also means that y^{2k+1}&gt;0. As a negative value for y would not satisfy this inequality, x>0 implies that y>0.

Furthermore, if -u = x, then:
x^{2k+1} = x*x^{2k} = -u*(-u)^{2k} = -u*u^{2k}=-u^{2k+1}
Therefore, proving x^{2k+1} = y^{2k+1} for x, y > 0 will also prove it for x, y < 0. Also, if x^{2k+1} = y^{2k+1} = 0, then:
x^{2k}*x = 0, y^{2k}*y = 0
x^{-2k} * x^{2k}*x = x^{-2k} *0, y^{-2k} * y^{2k}*y = x^{-2k} * 0
x=0, y=0
So, it suffices to focus on x,y>0.

x^{n} - y^{n} = (x-y)(x^{n-1} + x^{n-2}y + \ldots + xy^{n-n2} + y^{n-1}
Since x^{n} = y^{n}, x^{n} - y^{n} =0.
0 = (x-y)(x^{n-1} + x^{n-2}y + \ldots + xy^{n-n2} + y^{n-1}
Now, the second term must be positive, as it only contains multiplication and addition of positive terms. Thus, (x-y) = 0. Therefore x cannot be greater than or less than y, so x=y.

Sorry if that was a bit long; I was trying to make sure everything was explained completely.

 

[JonF]

i found your justification hard to follow here is an argument I would use.

x,y>0 is all that is needed because:

The x=0, y=0 case is trivial

if x ,y < 0 then let x= -a and y = -b for positive a,b. so xⁿ = yⁿ ⇔(-1)ⁿa ⁿ⇔ (-1)ⁿbⁿ ⇔a ⁿ=bⁿ since a,b are positive this means if x,y are negative we can generate an equivalent positive case.

If x^n = y^n and WLOG x>0 y<0 then let –a=y, where a is positive. So yⁿ = (-1)ⁿaⁿ = -(a)ⁿ. But we know xⁿ = yⁿ. This means xⁿ = -(a)ⁿ. But since x and a are both positive this means a positive equals a negative. This isn't possible so we know this case isn't possible.


Trichotomy covers why it is sufficient to show that both x<y, y>x aren't possible proves x=y. Since the cases are symmetric you only need to consider one WLOG.

 

[Strants]

Yeah, your argument is much easier to follow. I like the symmetry arguments; I hadn't thought of those! Thanks!

 

[╔(σ_σ)╝]

This seems like a straight forward induction proof. :-\ Try induction for a simple argument :-).

 

[hunt_mat]

Be wary of how you do the indunction argument for x^{n}=y^{n}, if you do the induction on n it will fail because if n in odd then n+1 will be even. Write n=2k+1 and try and induction argument on k.

 

[Strants]

OK, so n = 2k + 1
Let P(k) be x^2k+1 = y^2k+1 means x=y

P(0) is trivially true.

Assuming P(a), P(a+1) follows:
x^2a+1=y^2a+1, x=y
x²=y²
x^2a+1*x²=y^2a+1*y²
x^2(a+1)+1 = y^2(a+1)+1, x=y

Like that?

 

[╔(σ_σ)╝]

No, I think what he meant was write n=2k and perform induction on k.

Alternatively, you could do something funky like show that if n is true then 2n is true and also n-2 is true.

 

[Strants]

No, I think what he meant was write n=2k and perform induction on k.

But if I use n=2k, n would be even. I want to prove the statement for odd n, which would be 2k+1.

 

[╔(σ_σ)╝]

But if I use n=2k, n would be even. I want to prove the statement for odd n, which would be 2k+1.

I am sorry. You are correct. I thought you were supposed to prove it for even n.

What you did is correct.

 

[Strants]

I am sorry. You are correct. I thought you were supposed to prove it for even n.

That's fine; I need to do the proof for even n at some point, anyways.

I must admit, I'm a little curious; did you have the method to prove if n holds, then 2n and n-2 hold? It sounds like quite an interesting proof.

 

[╔(σ_σ)╝]

That's fine; I need to do the proof for even n at some point, anyways.

I must admit, I'm a little curious; did you have the method to prove if n holds, then 2n and n-2 hold? It sounds like quite an interesting proof.

First of all this is only true for n>= 2.

A sketch...

x^{k} = y^{k}

It is a simply matter to notice that squaring both sides preserves the inequality.

That is ...


x^{2k} = y^{2k}

The rest follows from the induction hypothesis.

 

[D H]

Hint: First explain why it suffices to consider only the case x, y >0; then show that x < y and x > y are both impossible.

The first part of the hint is essential. I don't see the value of the second.

This alone does it:

x^{n} - y^{n} = (x-y)(x^{n-1} + x^{n-2}y + \ldots + xy^{n-n2} + y^{n-1})

This means that

(a)\quad x-y = 0

(b)\quad x^{n-1} + x^{n-2}y + \ldots + xy^{n-n2} + y^{n-1} = 0

or (c), both (a) and (b) are true.

Now all that remains is showing that (b) is always positive when x,y>0.

 

[╔(σ_σ)╝]

The first part of the hint is essential. I don't see the value of the second.

This alone does it:

This means that

(a)\quad x-y = 0

(b)\quad x^{n-1} + x^{n-2}y + \ldots + xy^{n-n2} + y^{n-1} = 0

or (c), both (a) and (b) are true.

Now all that remains is showing that (b) is always positive when x,y>0.

The second hint eliminates all IN-equalities. :-)

It is more useful than the first hint.

 

[D H]

That's fine; I need to do the proof for even n at some point, anyways.

You can't. That x²ⁿ=y²ⁿ does not necessarily mean that x=y. Hint: x²=(-x)².

 

[╔(σ_σ)╝]

You can't. That x²ⁿ=y²ⁿ does not necessarily mean that x=y. Hint: x²=(-x)².

Not if you assume x,y>0; see original post.

 

[Strants]

You can't. That x²ⁿ=y²ⁿ does not necessarily mean that x=y. Hint: x²=(-x)².

Good point. The problem is to prove either x=y or x=-y.

Not if you assume x,y>0; see original post.

I was only able to assume that because n was odd: if x and y were of different signs, xⁿ and yⁿ would be, too. This does not hold for even n.

If I wanted to do an inductive proof that, for odd n, xⁿ = yⁿ implies x=y or x=-y, would this suffice?

Let n=2k
Let p(k) be the statement x^2k=y^2k means x=y or x=-y.
p(1) clearly holds (so does p(0), but I can't justify the inductive hypothesis without p(1)).
For p(k), p(k+1) also holds:
x^2k * x² = y^2k*y², x=y or x=-y
x^2(k+1)=y^2(k+1), x=y or x=-y

 

[D H]

OK, so n = 2k + 1
Let P(k) be x^2k+1 = y^2k+1 means x=y

P(0) is trivially true.

Assuming P(a), P(a+1) follows:
x^2a+1=y^2a+1, x=y
x²=y²
x^2a+1*x²=y^2a+1*y²
x^2(a+1)+1 = y^2(a+1)+1, x=y

Like that?

That proves that if x=y then x^2k+1=y^2k+1. It does not prove the converse.

Note that there is nothing in your proof that constrains the domain of x and y. In other words, were your proof valid, it would also work if x and y were complex. However, for complex x and y, that xⁿ=yⁿ does not mean that x=y, regardless of whether n is odd or even.

 

[Strants]

Almost missed your reply!

That proves that if x=y then x^2k+1=y^2k+1. It does not prove the converse.

I feel like you might be right, but my whole inductive hypothesis was if xⁿ=yⁿ and n is odd, then x=y. Given that, wouldn't I either have to have proved nothing (which is certainly possible), or have proved what I set out to prove?

Note that there is nothing in your proof that constrains the domain of x and y. In other words, were your proof valid, it would also work if x and y were complex. However, for complex x and y, that xⁿ=yⁿ does not mean that x=y, regardless of whether n is odd or even.

I though it was sort of implied that x and y were real, but you're right, that was sloppy of me. Just to make it explicit: x and y are real numbers.

 

[D H]

I feel like you might be right, but my whole inductive hypothesis was if xⁿ=yⁿ and n is odd, then x=y. Given that, wouldn't I either have to have proved nothing (which is certainly possible), or have proved what I set out to prove?

You proved exactly what I said you proved: If x=y then x^2k+1=y^2k+1. You did not prove the converse. There is nothing special about multiplying by x² on the left and y² on the right. You could just as well have multiplied by x and y. That leads to x=y ⇒ xⁿ=yⁿ. Just because you have proved that a implies b does not mean you have proved that b implies a.

 

[hunt_mat]

I just tried the proof by induction, it basically boils down to showing that x^2=y^2 then x=y, which would be true if x,y>0.

Mat

 

[D H]

Induction does not work here. Post #6 has a fatal flaw. It affirms the consequent, a formal fallacy. Just because A implies B does not mean that B implies A.

 

[Strants]

Looking over it again, D H is correct. P(0) is the assumption that x=y, and also the base case of the induction.

Thanks for the help, everyone! I appreciate it!

 

[╔(σ_σ)╝]

If the problem is proving the converse why don't you do so for odd n .?

In this case the contrapositive works well. It is easy to show that x =/= y implies x^n =/= y^n for odd n.

 

[Dickfore]

<br /> (1 + i \sqrt{3})^{3} = 1^{3} + 3 \cdot 1^{2} \cdot (i \sqrt{3}) + 3 \cdot 1 \cdot (i \sqrt{3})^{2} + (i \sqrt{3})^{3} = 1 + i 3 \sqrt{3} - 9 - i 3 \sqrt{3} = -8<br />

<br /> (-2)^{3} = -8<br />

 

[Char. Limit]

<br /> (1 + i \sqrt{3})^{3} = 1^{3} + 3 \cdot 1^{2} \cdot (i \sqrt{3}) + 3 \cdot 1 \cdot (i \sqrt{3})^{2} + (i \sqrt{3})^{3} = 1 + i 3 \sqrt{3} - 9 - i 3 \sqrt{3} = -8<br />

<br /> (-2)^{3} = -8<br />

I believe he meant to make the assumption that x and y are real.

 

[Strants]

If the problem is proving the converse why don't you do so for odd n .?

In this case the contrapositive works well. It is easy to show that x =/= y implies x^n =/= y^n for odd n.

But, would that also prove that x=y is the only possibility? I feel like it would leave the door open to the possibility that, say, x=-y (which is false, but just as an example).

 

[jgens]

But, would that also prove that x=y is the only possibility? I feel like it would leave the door open to the possibility that, say, x=-y (which is false, but just as an example).

Care to explain why you think that? If a statement is true, then it's contrapositive is also true. And the statement made by ╔(σ_σ)╝ is just the contrapositive of the original statement. I happen to disagree that it's any easier to prove than the original statement, but it's still true.

To address the concern you expressed above, if x=-y, then x != y so the proof would show that xⁿ != yⁿ which is the desired result.

 

[╔(σ_σ)╝]

Care to explain why you think that? If a statement is true, then it's contrapositive is also true. And the statement made by ╔(σ_σ)╝ is just the contrapositive of the original statement. I happen to disagree that it's any easier to prove than the original statement, but it's still true.

To address the concern you expressed above, if x=-y, then x != y so the proof would show that xⁿ != yⁿ which is the desired result.

I find it easier because if x! =y then x must either be less than y or greater than y (assume x, y are either both positive or negative for simplicity). In either case, x^n ! =y^n; a simple induction argument would show the contradiction.

 

[Strants]

To address the concern you expressed above, if x=-y, then x != y so the proof would show that xⁿ != yⁿ which is the desired result.

Yeah, I see that now. I guess I just wasn't thinking.

I find it easier because if x! =y then x must either be less than y or greater than y (assume x, y are either both positive or negative for simplicity). In either case, x^n ! =y^n; a simple induction argument would show the contradiction.

Couldn't I even do a proof by cases, if I didn't want to make that assumption? Something like this:

One may assume without loss of generality that x>y. In this case, there are three possibilities:

Case 0<y<x:
y²<x²
And I can multiply both sides of the inequality y<x by this to prove the statement for all odd n.

Case y<0<x,
Because y negative and y² is positive, yⁿ will be less that zero, and thus xⁿ, which will be positive for odd n.

Case y<x<0,
y² > x²
Because both sides of the inequality y<x are less than zero, multiplying the lesser side by a greater quantity than the greater side will preserve the inequality.