## Proof that irrational numbers cannot have repeating decimals.

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•••

= 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.