Question

Here's an easy question: prove that \[|x^n| = |x|^n \] for all positive n.

Answer 1

and for all real x

Answer 2

\[pmx ^{n}=(pmx)^{n}\]

Answer 3

isnt x^2 always positive to begin with?

Answer 4

oh, the original is x^n :)

Answer 5

Oh right hahaha.

Answer 6

when n is odd..

\[\left| x ^{n} \right| = -x ^{n} when x ^{n} < 0 , x <0 , \]

\[\left| x \right|^{n}= -x ^{n} , when, x <0 \]

Answer 7

the solution is only for when x<0 and n is odd...
because for even value of n and positive value of x solution can be understand by using same procedure...

Answer 8

try using induction

Answer 9

That's what I'm doing but I suck at induction...

Answer 10

Step one is to show that |x^n| = |x|^n for n = 1

Answer 11

Next step is to show that \[|x^k| = |x|^k , k=n \implies|x^{k+1}|=|x|^{k+1}\]

Answer 12

then show it is true for n+1 or not..
again you have to take the even and odd values of n...and positive and negative values of x...
same procedure...

Answer 13

why must we consider two different cases of even and odd n?

Answer 14

\[|x|=\sqrt{x^2}\]\[|x|^k=(\sqrt{x^2})^k=\sqrt{x^{2k}}=|x^k|\]\[|x|^{k+1}=(\sqrt{x^2})^{k+1}=\sqrt{x^{2k}}\sqrt{x^2}=\sqrt{x^{2k+2}}=\sqrt{x^{2(k+1)}}=|x^{k+1}|\]maybe...
I never know if I've done induction right :/

Answer 15

how is \( (\sqrt{x^2})^k = \sqrt{x^{2k}}\) ?

Answer 16

anyways, to continue with the inductive part, just say that \( |x|^{k+1} = |x| * |x|^k\) and proceed from there

Answer 17

again you stuck between even and odd..

Answer 18

\[(\sqrt{x^2})^n=(x^{2^{\frac{1}{2}}})^n\to x^{2*n*{\frac{1}{2}}}\to (x^{2n})^{\frac{1}{2}}\to\sqrt{x^{2n}}\]what's the matter with that?

Answer 19

Actually, I don't really see why induction is necessary in light of what I just showed.

Answer 20

\[|x|^n=(\sqrt{x^2})^n=(x^{2^{\frac{1}{2}}})^n\to x^{2*n*{\frac{1}{2}}}\to (x^{2n})^{\frac{1}{2}}\to\sqrt{x^{2n}}=|x^n|\]well that's backwards, but imagine it the other way ;-)