阅读下面材料:随着人们认识的不断深入,毕达哥拉斯学派逐渐承认
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmsqrt%3E%3Cmn%3E2%3C%2Fmn%3E%3C%2Fmsqrt%3E%3C%2Fmath%3E)
不是有理数,并给出了证明.假设是
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmsqrt%3E%3Cmn%3E2%3C%2Fmn%3E%3C%2Fmsqrt%3E%3C%2Fmath%3E)
有理数,那么存在两个互质的正整数p,q,使得
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmsqrt%3E%3Cmn%3E2%3C%2Fmn%3E%3C%2Fmsqrt%3E%3C%2Fmath%3E)
=
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmfrac%3E%3Cmi%3Ep%3C%2Fmi%3E%3Cmi%3Eq%3C%2Fmi%3E%3C%2Fmfrac%3E%3C%2Fmath%3E)
, 于是p=
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmsqrt%3E%3Cmn%3E2%3C%2Fmn%3E%3C%2Fmsqrt%3E%3C%2Fmath%3E)
q,两边平方得p
2=2q
2 . 因为2q
2是偶数,所以p
2是偶数,而只有偶数的平方才是偶数,所以p也是偶数.因此可设p=2s,代入上式,得4s
2=2q
2 , 即q
2=2s
2 , 所以q也是偶数,这样,p和q都是偶数,不互质,这与假设p,q互质矛盾,这个矛盾说明,
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmsqrt%3E%3Cmn%3E2%3C%2Fmn%3E%3C%2Fmsqrt%3E%3C%2Fmath%3E)
不能写成分数的形式,即
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmsqrt%3E%3Cmn%3E2%3C%2Fmn%3E%3C%2Fmsqrt%3E%3C%2Fmath%3E)
不是有理数.
请你有类似的方法,证明
不是有理数.