So far I have tried this:
Let $c = \gcd(a,b)$. We can say that $a = cx$ and $b = cy$ such that $x$ and $y$ are positive integers that are relatively prime. Then we have that $$ab = (cx)(cy) = c(xyc) = gcd(a,b)(xyc)$$ In order to show that $\gcd(, ) \cdot \mathrm{lcm}(, ) = ab$, we must show that $xyc = \mathrm{lcm}(a,b)$.
We will prove this by contradiction.
Assume that $z = \mathrm{lcm}(a,b)$ and that $z<xyc.$ We see that $xyc = (cx)y = ay$ and $xyc = (cy)x = bx$. This means that $a\mid xyc$ and $b\mid xyc.$ So, the lcm of $a$ and $b$ must also divide $xyc.$ Since $x\mid a$ and $y\mid b$, we know that $x\mid z$ and $y\mid z$. So, we can say that $xy\mid z.$
So z = xyn such that n is an integer and we will assume that $n<c$. Since z|xyc, we have that n|xyc. So c = nm where m is an integer greater than 1. So, we have: a = cx = nmx and z = xyn and that nmx|z so m|y b = cy = nmy and z = xyn and that nmy|z so m|x Since m > 1, we see that x and y are not relatively prime which means that $z<xyc$ cannot be true so z=xyc. So, we have that ab = cz = lcm(a,b) ⋅ gcd(a,b)
But I was told this is incorrect. Is there another approach?
