# Prove that $\lfloor \log_{10}(xy)\rfloor \geq \lfloor \log_{10}(x)\rfloor+ \lfloor \log_{10}(y)\rfloor$ [duplicate]

Let $x$ and $y$ be positive numbers. Prove that $$\lfloor \log_{10}(xy)\rfloor \geq \lfloor \log_{10}(x)\rfloor+ \lfloor \log_{10}(y)\rfloor.$$

I thought about using the definition of floor functions but didn't see how to use that here.

## marked as duplicate by Winther, Alex Mathers, Shailesh, Daniel W. Farlow, user91500Sep 22 '16 at 6:15

Since $\log_{10}(xy)=\log_{10}(x)+\log_{10}(y)$, it's enough to prove that $\lfloor a+b\rfloor \geq \lfloor a\rfloor +\lfloor b\rfloor$ for all real numbers $a,b$.
$\bigl\lfloor\log_{10}x\bigr\rfloor=k\iff 10^k\le x<10^{k+1}$.