Prove that there is an element in $ \{ \sqrt{3}, 2\sqrt{3}, 3\sqrt{3},...\}$ having fractional part less than 0.01 Given a set $ \{ \sqrt{3}, 2\sqrt{3}, 3\sqrt{3},...\}$, prove that some of the elements have fractional part less than 0.01 when written in decimal form.
Here is my attempt so far:
Divide the range $[0, 1.0]$ in blocks  of size 0.01. So there would be 100 such blocks. Now there blocks would act as our pigeon holes. Fractional part of subsequent set elements would act as our pigeons. Now either of two cases can happen:


*

*Fractional part of a set element lies in block $[0, 0.01]$. In this case we are done.

*2 set elements lie in same block. Let us say $n\sqrt{3}$ and $m\sqrt{3}$ where $n > m$. If mantissa of $n\sqrt(3)$ is greater than fractional part of $(n-m)\sqrt{3}$ would lie between $[0, 0.01]$. In this case we are done. I don't know what to do when fractional part of of $n\sqrt{3}$ is smaller. 


Any suggestions or alternate solutions? 

Based on original question and comments I will try to put together a complete proof:
Lema: Given a set $ \{ \sqrt{3}, 2\sqrt{3}, 3\sqrt{3},..., 101\sqrt{3}\}$, at least one of elements will have fractional part less than 0.01 or greater than 0.99.
Divide the range $[0, 1]$ in 100 cells of size 0.01. Assign all elements of the set to cells based on their fractional part. Since there are 101 elements in the given sequence, at least one cell will contain 2 elements. Lets say they are $p\sqrt{3}$ and $q\sqrt{3}$. Let also say that fractional part of $p\sqrt{3}$ is greater than fractional part of $q\sqrt{3}$. (Two fractional parts can not be equal, otherwise $\sqrt{3}$ would be rational number.)
If $p>q$, than $(p-q)\sqrt{3}$ will have fractional part less than 0.01, and will be one of elements of the initial set. And the element is found.
If $p<q$, than $(p-q)\sqrt{3}$ will have fractional part less than 0.01, but it will be negative number (lets say -2.998, -22.993, or similar), and $(q-p)\sqrt{3}$ will have fractional part greater than 0.99, and will be one of elements of the initial set. Then a procedure that Andre described is applied. And the element is again found.
