## proving that square root of 2 is irrational

Alright, I think I have a good way to prove that a natural number’s square root is either natural or irrational. Take for example 1, 4, or 9. Each of these number’s square root is a natural number (namely, 1, 2, 3). However, the square root of numbers such as 3, 5, 6, 7 and so on is not natural. We know that the square roots of those numbers are irrational, but just in case someone hopes that there is a natural number whose square root is a rational number, we will crush their hope proving that there is no such natural number.

As we often do with proofs, we assume the contrary to what we want to prove. We want to prove that there is no such natural number whose square root is rational. But let us assume that there is and let us name it M. So what then?! π

Not much, just that we assumed that there is a non-natural, rational number whose square equals M. Let us name this number Q.

Since Q is a rational number, we know that the number of digits after the decimal point is finite and that its last digit is not zero (any trailing zero after the last non zero digit can be ignored). As a result, we can safely assume that there is a multi digit natural number Nk and another natural number Na such that:

Q = Nk / (10^Na).

Also, we have assumed that Q^2 = M, which yields that Q^2 = Nk^2 / ((10^Na)^2) = M. This leads to:

Nk^2 = 10^(2Na)*M.

Since Nk’s last digit is non zero, Nk^2 ends with a non zero digit, and therefore is not divisible by 10. However, 10^(2Na)*M is divisible by 10 and equals Nk^2 (non divisible by 10), which is a contradiction. Therefore, we know that Q is neither natural nor rational, which means it must be irrational. π

Since 2 is a natural number, and its square root obviously is a non natural number, then it is irrational based on the above.

Hope anyone finds this of any help.

P.S. I had this in drafts since I was a student, before I started working, and I do not know why I did not publish it. I am publishing it now.