Is $|x| \cdot |x| = |x^2| = x^2$?

Question

Is $|x| \cdot |x| = |x^2| = x^2$ ?

I'm very sorry if this question is a duplicate but I couldn't find anything about it (most likely because it's wrong..). But I'm not sure if this is correct so I need to ask you.

$$|x| \cdot |x| = |x^2| \text{ should be alright}$$

Now my confusion starts. $x^2$ should be positive / neutral for any value. That would mean we can ignore the absolute value sign? On the other hand we could have that $|-x^2|$. But that would be a different thing than $|x^2|$, they are not equal to each other...? Please help me if I do this little thing wrong the entire task will be wrong. I got some thinking error here..

When there is the same question (I couldn't find one), please link me to it and I will delete this one immediately.

Answer 1

You are thinking it too hard. You could just look at the definition of the absolute value $$ |x|:=\begin{cases} x,&x\geq 0\\ -x,&x<0 \end{cases} $$ and check on your own that $|x|^2=|x^2|=x^2$.

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In general, we have $|a|\cdot|b|=|ab|$, which is true also for complex numbers; but the identity $|x^2|=x^2$ is not necessarily true in the complex world.

Answer 2

we have $$|x|\cdot |x|=|x|^2=x^2$$

Answer 3

For real numbers, this is true. We have:

$$|x| = sgn(x) \cdot x$$

And thus:

$$|x| \cdot |x| = sgn(x)^2 \cdot x^2 = x^2$$

Also:

$$|x^2| = sgn(x^2) \cdot x^2 = x^2$$

Answer 4

I think the confusion is coming from the fact that $|-a| = |a|$ is a perfectly valid statement. Absolute values, in a way, ignore the sign of a number. Also, you can ignore the absolute values signs in $|a|$ when you know that $a$ is always nonnegative. Thus $$ |x|^2 = |x| \cdot |x| = |x \cdot x| = |x^2| = x^2 $$ which could be seen in a variety of other ways.

P.S.: Don't feel bad about posting a possible duplicate question! It can be hard finding your question online, and so long as you tried to search for it, having your post flagged as a duplicate is not a bad thing.

Answer 5

The answer to your question is "yes". Since, as you point out, $x^2 \ge 0$, it is its own absolute value and you can "ignore the absolute value signs".

If $x \ne 0$ then $-x^2$ is negative, Its absolute value is $x^2$.

Answer 6

I think your confusion is confusing $-x^2 $ which is $-(x^2)$ which is always negative (or zero) and is not a square number. With $(-x)^2$ which is a square number and therefor always positive (or zero).

$-x^2 \ne (-x)^2$ (Unless $x = 0$).

Oh, and I guess you confusion is also $|-a| = |a|$. The absolute value of a negative value is not a different thing than an absolute value of a positive value. Indeed, the absolute value of a negative value is ALWAYS the exact same thing as the absolute value of the corresponding positive value. That is the entire point of absolute values.

So $|-5| = |5| = 5$ and $|-x^2| = |x^2| = x^2$.

And, for the record $|-x|*|-x| = |(-x)^2| = (-x)^2 = x^2$. While $|x|*|x| = |x^2| = x^2$ and ALWAYS $|-x| = |x|$. Always.

Answer 7

Simple: Look at the graph $y = x^2$; does $y$ ever submerge below the $y$-axis? That should give you an idea for all real numbers.

However, you have $a*i$ where $a$ is a real constant, and $i$ is an imaginary number. Squaring it gives $$(ai)^2 = a^2i^2 = a^2(-1) = -a^2$$ Therefore, if an imaginary number is squared, there is no positive solution.