## Why is the square root of ANY integer that is not a perfect square always an irrational number?

Posted by Greg in Special ProjectsIf, during those boring moments where you idly pressed numbers into an electronic calculator and randomly hit function buttons, you played extensively with the square root function – you might have seen something interesting. Provided that the integer you keyed in was not itself a perfect square, then the result of pressing the square root button always seemed to produce an irrational number – how odd is that then? I came across a proof that the square root of 2 is irrational – arrived at by a totally different route from the norm. The normal route is the centuries old reductio ad absurdum. This “new” route I found not only beautiful, but extremely powerful in that it proved that the square root of any integer that is not itself a perfect square will always yield an irrational. Here’s how it goes.

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.

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. Now the perfect square will decompose into ** pairs of primes** and this will give the possibility of even numbers of primes on both sides of the inequality and therefore the possibility of a rational solution.