How to prove that: let $n$ a natural odd, then prove that $x^n<y^n$ iff $x<y$

Question

I tried it in many ways but I couldn't prove it. TASK: Let $n$ a natural odd, $\ \ x,y\in\mathbb{R},$ then prove that $x^n<y^n$ iff $x<y$

My Attempt(s): $$\begin{align} & x^n<y^n \\ \Longleftrightarrow{} & x^n-y^n<0 \\ \Longleftrightarrow{} & (x-y)\sum_{k=0}^{n-1}x^{n-k-1}y^k<0 \end{align}$$ And I'm stuck here.

Thus, I can't solve it with induction either. Any help would be appreciated :)

Answer 1

What about considering function f(x) = x^n which is increasing function.

It means that for x < y -> f(x) < f(y).

So x^n < y^n

Answer 2

You are half-way there.

Your approach is clearly sufficient if $x$ and $y$ are both $> 0$, because then $\sum_{k=0}^{n-1}x^{n-k-1}y^k$ is strictly positive.

Now you need to show that if $x < 0$, then $x^n < 0$. Then you can just consider all the different cases of $x,y$ being strictly positive, strictly negative, and zero.

Answer 3

Hint: Start by proving the simpler, but easier version of the problem. If $0\leq x < y$, then $x^n<y^n$, for all natural numbers $n$. (You can easily prove this using induction.)

Hint: Realize that every number $x$ can be written as $x=\rm{sgn}(x) \cdot |x|$, where $\rm{sgn}(x)$ is the sign function. In particular, notice that if $n$ is odd, then $$\rm{sgn}(x)^n=\rm{sgn}(x)$$

You can now divide it into 4 cases:

$x<0,y<0$

$x<0,y\geq0$

$x\geq0,y<0$, but this case never happens since then $x>y$.

$x\geq0,y\geq0$, but you already proved this.

You can prove these two cases by writing $x=\rm{sgn}(x) \cdot |x|$ and $y=\rm{sgn}(y) \cdot |y|$, and using induction over $n$. Just be careful that now you need to prove that if the proposition is true for the odd natural $k$, then it is true for $k+2$ (which is the next odd natural number).

Answer 4

Hints:

By induction you can show that $x^n > 0$ holds for any $x>0$ and $(-1)^n = -1$ for odd $n$.

Prove the case $0\le x < y$ as you did.

The case $x \le 0 < y$ is trivial as $x^n = -(-x)^n$ is non positive and $y^n$ is positive.

The case $x < y \le 0$ use $$ -x^n = (-x)^n > (-y)^n = -y^n. $$

Fill the other cases.

Answer 5

It is better to prove that $f(x) = x^{n}$ (where $n$ is odd positive integer) is strictly increasing on both intervals $(-\infty, 0]$ and $[0, \infty)$ and this will imply that it is strictly increasing on whole of $\mathbb{R}$. The fact that $f(x)$ is strictly increasing on $[0, \infty)$ is obvious for the very reason that if $x, y \in [0, \infty)$ and $x < y$ then $y$ is strictly positive and $x$ is non-negative so that $$f(x) - f(y) = (x - y)\sum_{k = 0}^{n - 1}x^{n - 1 - k}y^{k} < 0$$ Let's check what happens on interval $(-\infty, 0]$. If $x, y \in (-\infty, 0]$ and $x < y$ then $x$ is strictly negative whereas $y$ is non-positive. Again we have $$f(x) - f(y) = (x - y)\sum_{k = 0}^{n - 1}x^{n - 1 - k}y^{k}$$ Note that the sum $\sum x^{n - 1 - k}y^{k}$ has terms where powers of $x, y$ have same parity (because their powers add up to $(n - 1)$ which is even). So if both the powers are even then the term is obviously non-negative. And if both the powers are odd then the term is product of two non-positive numbers and hence it is again non-negative. Thus each term in the sum $\sum x^{n - 1 - k}y^{k}$ is non-negative. Moreover first term $x^{n - 1}$ is positive because $x$ is negative and $n - 1$ is even. So the whole sum is positive and therefore $$f(x) - f(y) = (x - y)\sum_{k = 0}^{n - 1}x^{n - 1 - k}y^{k} < 0$$

Answer 6

It is elementary to show that $\sum_{k=0}^{n−1}x^{n−k−1}y^k\geq \frac{x^{n-1}+y^{n-1}}{2},$ provided $n$ is an odd natural number. With this, the desired result follows.