Assertion: Irrational numbers have decimal digits that do not repeat, nor do they have a pattern of digits that repeat.
An irrational number cannot be expressed as the ratio of two integers.
Proof by contradiction. Assume that the irrational number does have a repeating pattern of decimals.
First, consider an assumed irrational number x = a1a2•••am . am+1•••am+l•••x1x2•••xn•••x1x2•••xn•••
Where the digits a1a2•••am are the part of the number greater than one. Note the decimal point. The digits am+1•••am+l are a possible sequence of digits after the decimal that precede the repeating pattern. We will first show that we can eliminate these non repeating digits.
Multiply the number by 10m+l which gives
10m+l x = a1a2•••amam+1•••am+l . x1x2•••xn•••x1x2•••xn•••
= y + 0.x1x2•••xn•••x1x2•••xn•••
Where y is some whole number. We can rearrange
10m+l x - y = 0.x1x2•••xn•••x1x2•••xn•••
At this point we have reduced the problem to an assumed set of repeating decimals. We see that m or l could be 0, making for a simpler problem. Continuing, multiply by 10n
10n (10m+l x - y) = 10n 0.x1x2•••xn•••x1x2•••xn•••
= w + 0.x1x2•••xn•••
10m+l+n x - 10n y - w = 10lx - a1a2•••amam+1•••am+l
On the right we have rewritten 0.x1x2•••xn••• in terms of x. In other words, we have written just the repeating part in terms of x. Note that a1a2•••amam+1•••am+l is a whole number, call it u. Algebra gives
10m+l+n x - 10n y - w + u = 10lx
x = (10n y + w - u) / (10m+l+n - 10l)
Where the numerator and denominator on the right are both whole numbers, and therefore x is rational. This contradicts our assumption. To see this more simply, consider a case where x < 1 and does not have any decimals that are not part of the repeating series. This means m = l = y = 0, meaning all the a = 0.
Ok the super scripts don't work so I will try to fix this later.