



 
Hi Scarlet, It is the same sort of argument as is used to prove that the square root of 2, or the square root of 3, is irrational. Those are easy to find. Each is a proof by contradiction. You start by assuming that the square root of 3, say, is rational. Then there are integers m and n such that n is not zero and the square root of 3 is m/n. The fraction m/n can be assumed to be in lowest terms, therefore m and n have no common factor greater than 1. Squaring and rearranging gives you that 3n^{2} = m^{2}. Eventually, this implies that m is a multiple of 3. Taking advantage of this fact by writing m=3k, it follows after some rearranging and a similar argument that n is also a multiple of 3. This is a contradiction to the fact that m and n has no common factor greater than 1. Therefore the square root of 3 is irrational. Good luck.  


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