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## Homework Statement

Prove that there is no rational x such that x

^{2}=3

**2. The attempt at a solution**

Suppose that there is a rational x=[itex]\frac{a}{b}[/itex]=[itex]\sqrt{3}[/itex] and that the fraction is fully simplified. (ie. a and b have no common factor)

Then a

^{2}/b

^{2}=3 which means a

^{2}=b

^{2}.3 and it follows that a is a factor of 3 and be written a=3.k (k is an integer). Therefore a

^{2}/b

^{2}= 9.k

^{2}/b

^{2}=3. Therefore b

^{2}=3.k

^{2}.

This means both a and b are multiples of 3 and this contradicts our original assumption that the fraction is fully simplified.

**3. Question**

As far as I can tell the proof is sound but if you replace 3 with 4 the same logic holds and this means there is no rational number x such that x

^{2}=4 which is obviously wrong since 2 does. So what is wrong with this proof.