Proof of |x · y| = |x| · |y| using axioms of real numbers

Question

I'm trying to prove |x · y| = |x| · |y| using only the axioms of real numbers. I'm using the definition of the modulus function to be below. I thought I should start by distinguishing four cases like (x>0,y>0),(x<0,y<0),(y or x is equal to zero) and (y>0,x<0). Is this right? (https://i.stack.imgur.com/1qHgQ.gif)

Answer 1

Hint for an alternative proof without using cases.

• Prove that $\;\;|a|^2=a^2$

• Prove that $\;\;|a|=|b| \;\; \iff \;\; a^2=b^2$

• Prove that $\;\;(x \cdot y)^2 = x^2 \cdot y^2$

• Deduce from the above that $\;\;|x \cdot y|=|x| \cdot |y|$