Problem #PRU-100258

Problem

Suppose $a\ne b$. We can write $$-a=b-(a+b)$$ and $$-b=a-(a+b)$$ Since $(-a)b=a(-b)$, then $$(b-(a+b))b=a(a-(a+b))$$ Removing the brackets, we have $$b^2-(a+b)b=a^2-a(a+b)$$ Adding ${\left(\frac{a+b}{2}\right)}^2$ to each member of the equality we may complete the square of the differences of two numbers $${(b-\frac{a+b}{2})}^2={(a-\frac{a+b}{2})}^2$$ From the equality of the squares we conclude the equality of the bases $$b-\frac{a+b}{2}=a-\frac{a+b}{2}.$$ Adding $\frac{a+b}{2}$ to both sides of equality we get $a=b$. Therefore, WE HAVE SHOWED THAT FROM $a\ne b$ IT FOLLOWS $a=b$.