What does $|a| = |b|$ is equal to?

Question

I want to know what does $|a| = |b$| is equal to. I think it's equal to $a = \pm b$. Please tell me if I'm right or if I'm wrong.

Answer 1

It is equivalent to:

$a=b$ if both are positive or both are negative, or both are zero, and

$a=-b$ if their signs are different.

It means their magnitudes are the same, but not necessarily their signs.

It's important to note that if you restrict yourself to the positive integers then this means $a=b$ as there is no number $-a$ or $-b$ so $\lvert a\rvert = a $ but in the most common high school setting of all real numbers it is as above.

The statement has a more complex meaning in the complex numbers $x=re^{i\theta}$ in which case $\lvert x\rvert=r$, which is again a measure of magnitude with the measure of positive or negative abstracted away. But in this case rather than the simple $\pm$ being abstracted away, we are removing a more sophisticated measure of the "positiveness" which is $\theta$; a direction vector whose proximity to $\pi$ or $2\pi$ in a sense measures how positive or negative a number is on a circular scale.

Answer 2

I think it's better to say that it's equivalent to $$a=b$$ or $$a=-b.$$

Answer 3

It depends on what you're dealing with. If you're dealing with real numbers, then you're correct: either $a=b$ or $a=-b$ (or both, if $a=b=0$).

If you're dealing with complex numbers, the best you can say is that $a=be^{i\theta}$ for some real $\theta.$

Answer 4

Consider, $$ a^2=b^2$$ which is the same as $$|a|=|b|$$ (Removing even powers results in modulus).

Now $$a^2= b^2 \iff (-a)^2 = a^2 = b^2 = (-b)^2 $$

And that's why, $$|a|=|b| \iff |a|=±b$$ but $$|a|=-b $$ is not possible as modulus of any thing cannot be negative. So $$|a| = b$$

$$ \iff ±a=b $$

$$ \text{OR} \ \ \ \ \ a=b , a=-b \ \ \ \ \ \square $$

I would like to give $ \text{ONE MORE} $ proof.

We have, $$|a|=|b|$$

So, consider the most simple implication, $$ |a| = ±b \ \ \ \ \text{AND} \ \ \ \ ±a =|b| $$ See, we've used 'AND' but not 'OR' , which means the intersection of both cases.

Proceeding further,

$$ |a|= b \ \ \ \ \text{AND} \ \ \ \ a =|b| $$ $$ ±a = b \ \ \ \ \text{AND} \ \ \ \ a =±b $$ $$ ( a = b \ OR \ -a = b ) \ \ \ \ \text{AND} \ \ \ \ ( a = b \ OR \ -b ) $$ $$ \text{COMBINING BOTH} \ \ \ \ a=b, a=-b$$