Problem 6

(a) Prove that if $0 \leq x < y$, then $x^n < y^n$, $n = 1, 2, 3, ...$.

We can prove that by induction. $n = 1$: \[ \begin{align*} x &< y && \text{| which is obviously correct} \\ \end{align*} \] $n = 2$: \[ \begin{align*} x^2 &< y^2 && \text{| which is true based on 5.(ix)} \\ \end{align*} \] $n = k$ is true, show that $n = k + 1$ is also true: \[ \begin{align*} x^{k + 1} &< y^{k + 1} \\ xx^{k} &< yy^{k} && \text{| based on 5.(viii) with } a &= x, b &= y \\ x^k &< y^k \\ \end{align*} \]

(b) Prove that if $x < y$ and $n$ is odd, then $x^n < y^n$.

For $0 \leq x < y$ this is true based on (a). Now we have to show that it also holds for (A) $x < 0, y \geq 0$ and (B) $x, y < 0$.

Starting with (A) using induction - $n = 1$: \[ \begin{align*} x &< y \\ \end{align*} \] $n = 3$: \[ \begin{align*} x^3 &< y^3 && \text{| with } x = (-1)a, a = |x| \\ (-1)(-1)(-1)a^3 &< y^3 \\ (-1)|x|^3 &< y^3 && \text{| which is true because } y \geq 0 \rightarrow y^n \geq 0 \\ \end{align*} \] $n = k$ is true with $k$ odd, show that $n = k + 2$ is also true: \[ \begin{align*} x^{k + 2} &< y^{k + 2} \\ (-1)^{k + 2}a^{k + 2} &< y^{k + 2} && \text{| with } x = (-1)a, a = |x| \\ (-1)a (-1)^{k + 1}a^{k + 1} &< yy^{k + 1} && \text{| using 5.(viii) with } x < y \\ (-1) (-1)^{k + 1}a^{k + 1} &< y^{k + 1} \\ (-1)(-1) (-1)^{k}a^{k + 1} &< y^{k} \\ (-1)(-1) x^k &< y^k \\ x^k &< y^k \\ \end{align*} \] (B) is identical to (a) when we just remove $-1$ on both sides and continue with $|x|$ and $|y|$.

(c) Prove that if $x^n = y^n$ and $n$ is odd, then $x = y$.

For $x, y > 0$ this statement holds true for any $n$. When $x, y < 0$ we can remove $-1$ again on both sides and deal with $|x|$ and $|y|$ which is true for any $n$. What is left is to show that (A) $x < 0, y > 0$ and (B) the inverse of (A) are only possible for odd $n$ when we remove the sign. Using induction - $n = 1$: \[ \begin{align*} (-1)x &= y \\ (-1)(-1)|x| &= y \\ |x| &= y && \text{| otherwise it is a contradiction} \\ \end{align*} \] Using induction for (A) - $n = 3$: \[ \begin{align*} (-1)x^3 &= y^3 \\ (-1)(-1)(-1)(-1)|x|^3 &= y^3 \\ |x|^3 &= y^3 \\ \end{align*} \] $n = k$ is true with $k$ odd, show that $n = k + 2$ is also true: \[ \begin{align*} (-1)x^{k + 2} &= y^{k + 2} \\ (-1)(-1)(-1)(-1)^k|x|^k &= y^k \\ (-1)(-1)^k|x|^k &= y^k \\ (-1)x^k &= y^k \\ \end{align*} \] This proof by indiction works exactly the same for (B).

(d) Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$.

In case of $x, y > 0$ and $x, y < 0$ the same arguments applies as for (c) so we have to show again that $x > 0, y < 0$ is possible for even $n$.

Using induction for (A) - $n = 2$: \[ \begin{align*} x^2 &= y^2 \\ x^2 &= (-1)(-1)|y|^2 \\ x^2 &= |y|^2 \\ \end{align*} \] $n = 4$: \[ \begin{align*} x^4 &= y^4 \\ x^4 &= (-1)(-1)(-1)(-1)|y|^4 \\ x^4 &= |y|^4 \\ \end{align*} \] $n = k$ is true with $k$ even, show that $n = k + 2$ is also true: \[ \begin{align*} x^{k + 2} &= y^{k + 2} \\ x^k &= (-1)(-1)(-1)^k|y|^k \\ x^k &= (-1)^k|y|^k \\ x^k &= y^k \\ \end{align*} \]

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