Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.



Re: Reductio Ad Absurdum
Posted:
Jan 16, 1998 12:01 PM


On Fri, 16 Jan 1998, William wrote:
> Dear Sir, > please help me prove that the square root of 2 is an irrational > number using "reductio ad absurdum". Please explain the method. > > Thanks a million > 
William,
This is not the usual proof that you can find in many math books. But I like it.
If the square root of two were not irrational then it could be written as a fraction a/b where a and b are both whole numbers. Since fractions can be reduced we may as well assume that a and b have no common factor. If a and b have no common factor then a^2 and b^2 cannot have a common factor either.
Square both sides to get a^2 / b^2 = 2 and multiply through by b^2 to get a^2 = 2 ( b^2)
In decimal notation the only possible last digits of a and b are 0,1,2,3,4,5,6,7,8, or 9. The last digits of a^2 and b^2 can only be 0, 1, 4, 9, 6, or 5.
And 2(b^2) must end in 0, 2, or 8.
Since a^2 = 2(b^2) the only possibility is that both end in 0. So a^2 ends in zero and b^2 ends in either 0 or 5. In either case 5 is a common factor of a^2 and b^2 . . . a contradiction.
Alan
 Alan Lipp Williston Northampton School Easthampton, MA



