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Langara College - Department of Mathematics and Statistics
Internet Resources for the Calculus Student - Topics in Precalculus
The Irrationality of Root 2
Any ratio of integers can be reduced to "lowest terms" by cancelling out
common factors.
So if root 2 is rational, then it can be expressed as a ratio
m/n where m and n have no common factors.
Now if m/n is root 2, then (m^2)/(n^2)=2, so m^2 = 2*(n^2) is even.
But the square of any odd number is of the form (2k+1)^2=4k^2+4k+1=even+1, which is odd.
So, since we just showed that m^2 is even, m must be even also - ie m=2k for some whole number k.
Thus m^2 =(2k)^2=4k^2 is actually a multiple of 4.
But then 2*(n^2)=(m^2)=4*something, so (n^2)=2*something, and so it
too is even.
So n must be even also.
But then m and n both have a factor of two and we already cancelled
out all common factors!
Thus if we assume that root 2 is a ratio of whole numbers then we
get a contradiction,
and so it cannot be true.
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It is also true that the square root of any integer that is not a perfect square is irrational
A number of other ways of proving this have been collected on Alex Bogomolny's 'Cut-the-Knot' website.