|xⁿ| ≡ |x|ⁿ ≡ xⁿ

[burgerandcheese]

Hi. I was learning modulus functions when I came across this:

"If x is any real number, then |x²| ≡ |x|² ≡ x²."

I was wondering if we could generalise it even further to:

If x is any real number, then |xⁿ| ≡ |x|ⁿ ≡ xⁿ for even n.

I asked two of my math teachers and on a Facebook group but the responses were:

1. |xⁿ| ≡ |x|ⁿ ≡ xⁿ if x≥0 and (-x)ⁿ if x<0 for any n

2. Yes

3. No such thing

What is correct?

 

[Romsek]

\(\displaystyle \text{If $n$ is even, then $|x^n| = |x|^n,~\forall x \in \mathbb{R}$}\)

\(\displaystyle \text{If $n$ is odd, then $\begin{cases}|x^n| = |x|^n &x\geq 0\\|x^n| = -|x|^n &x < 0\end{cases}$}\)

 

[pka]

\(\displaystyle \text{If $n$ is even, then $|x^n| = |x|^n,~\forall x \in \mathbb{R}$}\)
\(\displaystyle \text{If $n$ is odd, then $\begin{cases}|x^n| = |x|^n &x\geq 0\\|x^n| = -|x|^n &x < 0\end{cases}$}\)

It is never true that if \(\displaystyle x\ne 0\) then \(\displaystyle |x^n|=-|x|^n\)

"If x is any real number, then |x²| ≡ |x|² ≡ x²." CORRECT
I was wondering if we could generalise it even further to:
If x is any real number, then |xⁿ| ≡ |x|ⁿ ≡ xⁿ for even n. Correct

First of all we are using real numbers. In general \(\displaystyle n\in\mathbb{Z}^+\)
Ask yourself if \(\displaystyle |(-5)^3|=|-5|^3\) Well yes of course!
Then \(\displaystyle \forall n[|x^n|=|x|^n]\) that is not in dispute!
BUT does \(\displaystyle |(-5)^3|=(-5)^3~?\) well of course not! Because \(\displaystyle \forall x[|x|\ge 0]\).
It is true that if \(\displaystyle n\) is even it is true that \(\displaystyle |x^n|=(x)^n\)