Help on proof by contraposition

How do I prove $\displaystyle{x^n<y^n \Leftrightarrow x<y}$ using proof by contraposition?

Any ideas? Thanks

Wait, I meant how would I go about it for for $\displaystyle{x^n<y^n \Rightarrow x<y}$? Sorry for the hassle.

I read contraposition as constipation
sorry I have absolutely nothing useful to contribute to this thread.

So not(Q) is $\displaystyle{y<x}$ and not(P) is $\displaystyle{y^n<x^n}$?

So then can I say not(Q) implies not(P) by induction?

What other way can i prove it? :s

I read contraposition as constipation
sorry I have absolutely nothing useful to contribute to this thread.

At some point you will have to end up proving that f(x) = x^n is an increasing function for certain conditions on x and n.

So, if P implies Q, which was shown by the above line, lets focus on P and specifically not(P).
If, not(P), namely, x^n can be bigger than or equal to y, then we have the following:

x^n >= y^n implies x >= y, just by taking the Nth root as above, and namely not(P) implies Z, where the statement Z is x >= y.
This is in contradiction with the above and not(Z) implies Q. (Z is actually not(Q)).
Therefore, for the statement Q to hold, we require that x is less than y.

So can I just say:
If $\displaystyle{x^n>=y^n \Rightarrow x>=y}$ by taking the nth root, then not(P) implies not(Q) by contraposition and thats the proof?

Also, I've assumed that x and y are positive real numbers and n is a natural number.

I would say yes.
However, I will try to explain all the steps by formalising them a little and to give another view on the problem.

The statement which is to be proved is:
$x^n < y^n \Rightarrow x < y$

Therefore, assuming x,y are real numbers and n is a natural, you can show that this holds by taking the Nth root.
Nevertheless, this won't be a proof, since the direction of the logic is not right, as I said before.
(You have inverted it in your latest post, which is the right way)

To clarify the point further, I will borrow one example from Martin Liebeck's Introduction to Pure Mathematics:

Prove that $\sqrt 2 + \sqrt 6 < \sqrt 15$

1. Non-proof of the statement

$[br]\sqrt 2 + \sqrt 6 < \sqrt 15 \Rightarrow (\sqrt 2 + \sqrt 6)^2 < 15[br]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow 8 + 2\sqrt 12 < 15 \Rightarrow 2\sqrt 12 < 7 \Rightarrow 48 < 49[br]$

This is true of course, but here is Liebeck's comment on it:
"The last statement (48 < 49) is true, so why is this not a proof? Because the implication is going the wrong way - we have shown that if $P$ is the statement we want to prove, and $Q$ is the statement that 48 < 49, then $P \Rightarrow Q$; but this tells us nothing about the truth or otherwise of $P$.
A cunning change to the above false proof gives a correct proof, by contradiction. So assume the result is false; i.e., assume that $\sqrt 2 + \sqrt 6 \geq \sqrt 15$. Then ..."

2. Proof by contradiction

$[br]\sqrt 2 + \sqrt 6 \geq \sqrt 15 \Rightarrow (\sqrt 2 + \sqrt 6)^2 \geq 15[br]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow 8 + 2\sqrt 12 \geq 15 \Rightarrow 2\sqrt 12 \geq 7 \Rightarrow 48 \geq 49[br]$

".. which is a contradiction."


O.K., how is that related to your problem?
Your statement $P$, or $x^n < y^n$, can be shown as $x < y$ by taking the Nth root, but this will be non-proof as the above.
So, to prove the statement by contradiction (contraposition) you should assume the opposite; i.e., for all $x^n$ and $y^n$, if $x^n \geq y^n$, then by taking the Nth root you can show that $x \geq y$, and therefore to prove your original statement.
(As what we have shown and you adapted as a proof, so far)



**
Here is the formalised statement and a "proper" proof by contradiction:

This first statement (1) is the equivalent of your (edited) statement:

(1) $\forall x,y,n \in \{k \ | \ k \in \mathbb{R}, k \geq 0 \}$ such that $x^n < y^n, \ \ x < y$;
-
. For all $x$, $y$ that are positive real numbers and such that $x^n < y^n$, the inequality $x < y$ will be true for all positive real values of $n$.
. This is in regard to $x, y$ and $n$ being positive real numbers.

To prove that, I will take its negation and attempt to disprove it:

N(1) $\exists x,y \in \{k \ | \ k \in \mathbb{R}, k \geq 0 \}$ such that $x \geq y, \ \ x^n < y^n$ for some $n \in \{k \ | \ k \in \mathbb{R}, k > 0 \}$
-
. There exist positive real numbers $x$ and $y$, such that if $x \geq y$, the inequality $x^n < y^n$ is true for some positive real number $n$ greater than 0.

Therefore, exponentiating both sides by a positive real number won't change the inequality and:

$x \geq y \Rightarrow x^n \geq y^n$, which is a contradiction and it follows that N(1) is false.



***
Now just for fun, I will prove by contradiction a special case for odd powers of x and y.

(2) $\forall x,y \in \mathbb{R} \$ such that $x^n < y^n,\$ and the inequality $\ x < y\$ is true $\ \forall n \in \{k \ | \ k \in \mathbb{N}, \ 2k + 1\}$

Its negation:

N(2) $\exists x,y \in \mathbb{R} \$ such that $x \geq y,\$ and the inequality $\ x^n < y^n\$ is true for some $\ n \in \{k \ | \ k \in \mathbb{Z}, \ k \geq 0, \ 2k + 1\}$

Hence, exponentiating both sides by an odd natural power:

$x \geq y \Rightarrow x^{(2n + 1)} \geq y^{(2n + 1)} \Rightarrow x \times x^{(2n)} \geq y \times y^{(2n)}$, which is a contradiction and it follows that N(2) is false.



If you combine (1) and (2), you have the most of your original statement; it shows all positive plus a special case for the negatives.
And it is certainly not true in general, for example: -3 < -2, but (-3)^2 is bigger than (-2)^2.
(However, if you wan't to stick with the Nth root, you should bear in mind the irrationals and zeros)



It is also possible to express a proof by using the properties of the derivatives of odd and even functions, as suggested by @around.
I don't think it would be easier to prove it that way, but if you need to show this by using calculus, this is the way.

I would say yes.
However, I will try to explain all the steps by formalising them a little and to give another view on the problem.

The statement which is to be proved is:
$x^n < y^n \Rightarrow x < y$

Therefore, assuming x,y are real numbers and n is a natural, you can show that this holds by taking the Nth root.
Nevertheless, this won't be a proof, since the direction of the logic is not right, as I said before.
(You have inverted it in your latest post, which is the right way)

To clarify the point further, I will borrow one example from Martin Liebeck's Introduction to Pure Mathematics:

Prove that $\sqrt 2 + \sqrt 6 < \sqrt 15$

1. Non-proof of the statement

$[br]\sqrt 2 + \sqrt 6 < \sqrt 15 \Rightarrow (\sqrt 2 + \sqrt 6)^2 < 15[br]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow 8 + 2\sqrt 12 < 15 \Rightarrow 2\sqrt 12 < 7 \Rightarrow 48 < 49[br]$

This is true of course, but here is Liebeck's comment on it:
"The last statement (48 < 49) is true, so why is this not a proof? Because the implication is going the wrong way - we have shown that if $P$ is the statement we want to prove, and $Q$ is the statement that 48 < 49, then $P \Rightarrow Q$; but this tells us nothing about the truth or otherwise of $P$.
A cunning change to the above false proof gives a correct proof, by contradiction. So assume the result is false; i.e., assume that $\sqrt 2 + \sqrt 6 \geq \sqrt 15$. Then ..."

2. Proof by contradiction

$[br]\sqrt 2 + \sqrt 6 \geq \sqrt 15 \Rightarrow (\sqrt 2 + \sqrt 6)^2 \geq 15[br]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow 8 + 2\sqrt 12 \geq 15 \Rightarrow 2\sqrt 12 \geq 7 \Rightarrow 48 \geq 49[br]$

".. which is a contradiction."


O.K., how is that related to your problem?
Your statement $P$, or $x^n < y^n$, can be shown as $x < y$ by taking the Nth root, but this will be non-proof as the above.
So, to prove the statement by contradiction (contraposition) you should assume the opposite; i.e., for all $x^n$ and $y^n$, if $x^n \geq y^n$, then by taking the Nth root you can show that $x \geq y$, and therefore to prove your original statement.
(As what we have shown and you adapted as a proof, so far)



**
Here is the formalised statement and a "proper" proof by contradiction:

This first statement (1) is the equivalent of your (edited) statement:

(1) $\forall x,y,n \in \{k \ | \ k \in \mathbb{R}, k \geq 0 \}$ such that $x^n < y^n, \ \ x < y$;
-
. For all $x$, $y$ that are positive real numbers and such that $x^n < y^n$, the inequality $x < y$ will be true for all positive real values of $n$.
. This is in regard to $x, y$ and $n$ being positive real numbers.

To prove that, I will take its negation and attempt to disprove it:

N(1) $\exists x,y \in \{k \ | \ k \in \mathbb{R}, k \geq 0 \}$ such that $x \geq y, \ \ x^n < y^n$ for some $n \in \{k \ | \ k \in \mathbb{R}, k > 0 \}$
-
. There exist positive real numbers $x$ and $y$, such that if $x \geq y$, the inequality $x^n < y^n$ is true for some positive real number $n$ greater than 0.

Therefore, exponentiating both sides by a positive real number won't change the inequality and:

$x \geq y \Rightarrow x^n \geq y^n$, which is a contradiction and it follows that N(1) is false.



***
Now just for fun, I will prove by contradiction a special case for odd powers of x and y.

(2) $\forall x,y \in \mathbb{R} \$ such that $x^n < y^n,\$ and the inequality $\ x < y\$ is true $\ \forall n \in \{k \ | \ k \in \mathbb{N}, \ 2k + 1\}$

Its negation:

N(2) $\exists x,y \in \mathbb{R} \$ such that $x \geq y,\$ and the inequality $\ x^n < y^n\$ is true for some $\ n \in \{k \ | \ k \in \mathbb{Z}, \ k \geq 0, \ 2k + 1\}$

Hence, exponentiating both sides by an odd natural power:

$x \geq y \Rightarrow x^{(2n + 1)} \geq y^{(2n + 1)} \Rightarrow x \times x^{(2n)} \geq y \times y^{(2n)}$, which is a contradiction and it follows that N(2) is false.



If you combine (1) and (2), you have the most of your original statement; it shows all positive plus a special case for the negatives.
And it is certainly not true in general, for example: -3 < -2, but (-3)^2 is bigger than (-2)^2.
(However, if you wan't to stick with the Nth root, you should bear in mind the irrationals and zeros)



It is also possible to express a proof by using the properties of the derivatives of odd and even functions, as suggested by @around.
I don't think it would be easier to prove it that way, but if you need to show this by using calculus, this is the way.