If you have played with the square root function on an electronic calculator you may have noticed something odd. It looks like if you take the square root of * ANY *number that is not actually a perfect square, then you get an irrational number, with the decimal places spilling over the edge of your calculator’s display. Is this in fact true? Is the square root of any number that is not itself a perfect square an irrational? Let’s find out.

We start off in the conventional way by assuming that the square root of 2 * IS* rational and may be given as m/n where m & n are both integer.

1) √2 = m/n

2) m^{2} = 2 n^{2}

That much we have seen before, but now comes the really clever bit. Factor m & n into their unique prime factors:

3) m_{1} m_{1} m_{2} m_{2 }m_{3} m_{3} … m_{r} m_{r} = 2 n_{1} n_{1} n_{2} n_{2} n_{3} n_{3} … n_{r} n_{r}

Now here’s the problem. The prime number 2 will appear an even number of times on the left hand side of the equation (if it appears at all) and an odd number of times on the right. Since the decomposition into primes is unique the prime number 2 cannot appear an even number of times on one side of the equality and an odd number of times on the other. So the square root of 2 cannot be written in the form m/n with m & n integer. The square root of 2 is an **irrational number.**

A) Although the above was shown to be true for root 2, the same argument of course would hold true for *any prime number. The square root of any prime number is an irrational.*

B) Replace 2 with ** any** integer that is not a perfect square. Now that integer may be decomposed into its prime factors, and, if the number is not a perfect square, then once again we will have an odd number of primes on one side of the equality and an even number of primes on the other.

*The square root of any integer that is not a perfect square is also an irrational.*C) Finally, replace 2 with any perfect square. With even numbers of primes on both sides the solution is trivial.