Proving that if $x,y \in \Bbb R$ and $|x|=|y|$ then $x^2=y^2$

Question

I've seen that if $|x|=|y|$ for reals $x$ and $y$ that $x^2$ is also equal to $y^2$. Generally that makes handling the absolute values much easier algebraically. I was wondering if there was any more need proof beyond If $x,y \in \Bbb R$ then $|x|=\sqrt{x^2}$ so if $|x|=|y|$ then $\sqrt{x^2}=\sqrt{y^2}$ square both sides: $x^2=y^2$.

I was also wondering if it would be valid to say if $x,y \in \Bbb R$ and $|x|=y$ then $x=\pm y$ and $\pm x=y$, and in extension $x^2=y$ then $x=\pm \sqrt{y}$.

Answer 1

Assuming it is known that:

$\;(1)\quad\quad |a| \cdot |b| = | a \cdot b|$

$\;(2)\quad\quad a^2 \ge 0$ for $\forall a \in \mathbb{R}$

$\;(3)\quad\quad a \ge 0 \implies |a| = a$

The proof follows in one line: $$x^2 \stackrel{(2)+(3)}{\;=\;} |x^2| = |x \cdot x| \stackrel{(1)}{\,=\,} |x| \cdot |x| = |x|^2 = |y|^2 = |y|\cdot|y| \stackrel{(1)}{\,=\,} |y \cdot y| = |y^2| \stackrel{(2)+(3)}{\;=\;} y^2$$

Answer 2

Just use that

$$\vert x\vert^2=x^2$$

In fact the last equality comes from $$(\vert x\vert -x)(\vert x\vert+x)=0$$ because one of the two factors is $0$ using the definition of the absolute value.

Answer 3

By definition of absolute value, $x=\pm y$.

If $x=y$, then $$x^2=x\cdot x=y\cdot y=y^2.$$ If $x=-y,$ then $$x^2=x\cdot x=(-y)\cdot(-y)=y\cdot y=y^2$$

Answer 4

Thats a good question.

I think it is more consequential than direct though.

For real numbers,

We have as an axiom for $a > 0$ and $x < y$ than $ax < ay$. This gives us: if $x > 0$ then $x^2 = x*x > x*0 = 0$.

We also have as an axiom that for ever real $b$ there is a unique $-b$ so that $b + (-b) = 0$. And we have an axiom that if $x < y$ than $x + c < y +c$ for any $c$. From this we have $b > 0 \iff b +(-b) > 0 + (-b) \iff 0 > -b$.

And there therefore we have a proposition that for $a < 0$ and $x < y$ than $ax > ay$. (Because $a < 0 \implies -a > 0 \implies -ax < -ay \implies ax > ay$.) [Okay, I skipped proving that $(a)(-b) = -(ab)$ but... that's details...].

Thus we have. For any $x \in \mathbb R$ exactly one of the following are true:

i) $x > 0$ and $x^2 > 0$.

ii) $x = 0$ and $x^2 = 0$.

iii) $x < 0$ and $x^2 = x*x > x*0 = 0$.

So all squares are positive or zero (and only zero if it is zero squared).

.... Hmm.... okay, we do need to prove that $(a)(-b) = -(ab)$ and that $-(-a) = a$ and that $(-a)(-b) = ab$, after all.

We have an axiom that $a(b+c) = ab+bc$. So $ab+ (a)(-b) = a(b + (-b)) = a*0 = 0$. Now we have an axiom that there is one unique $-ab$ so that $ab + (-ab) =0$ and $ab + (a)(-b) =0$.... so that mean $(a)(-b)$ must be equal to that unique $-ab$.

We also have $a + (-a) = -a + a = 0$ so there is a unique $-(-a)$ so that $-a + (-(-a)) = 0$. So that must be $-(-a) = a$.

And finally $(-a)(-b) = -(-a)(b) = --(ab) = ab$.

So .... given all that....

AND defining that $|a| = a$ if $a > 0$ and $|a| = 0$ if $a = 0$ and that $|a| = -a$ (which is positive) if $a < 0$.

We can have:

$|a| = |b|$ means either.

1) $a = b$ and $a\ge 0 ; b\ge 0$ or $a < 0; b < 0$ and $a^2 = b^2$.

2) $a = -b$ and either $a \ge 0; b \le 0$ or $a < 0; b> 0$ and then $a^2 = (-b)^2 = b^2$.