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Question 281872: prove that square root of 3 is irrational
Answer by nabla(475)   (Show Source):
You can put this solution on YOUR website! Assume to the contrary that sqrt(3) is rational. That is, it can be expressed as sqrt(3)=x which is the same as
(1) 3=x^2
Now, let x=m/n where gcd(m,n)=1.
3=m^2/n^2 follows from (1). Through some manipulation we find
3n^2=m^2. This means that 3 divides m^2 (in symbols 3|m^2) moreover by properties of squares 3|m. So, we can represent m as
(2) m=3k.
By (2), 3n^2=(3k)^2=9k^2
which gives n^2=3k^2 so 3|n^2 which as above means 3|n. However, we assumed that gcd(m,n)=1 (i.e. that it is a simplified rational!). Obviously gcd(m,n)>=3.
So we have a contradiction and therefore sqrt(3) is not a rational number hence it is an irrational number which is precisely R\Q.
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