有5个网友回答
1、假设根号3=p/q(p、q为互质整数),则p^2=3q^2
所以3整除p^2,因3是质数,所以3整除p,可设p=3t,则q^2=3t^2,所以3整除q
因此p和q有公约数3,与p和q互质矛盾,所以根号3是无理数
2、设x=根号3,则有方程x^2=3
假设x^2=3有有理数解x=p/q(p、q为互质整数),根据牛顿有理根定理p整除3,q整除1,所以p=1或3,q=1,从而x=1或3,显然x=1或3不是方程x^2=3的根,矛盾.
3、设x=根号3=p/q,(p,q)=1,所以存在整数s,t使ps+qt=1
根号3=根号3*1=根号3(ps+qt)=(√3p)s+(√3q)t=3qs+pt为整数,矛盾

拓展资料:
由无理数引发的数学危机一直延续到19世纪下半叶。1872年,德国数学家戴德金从连续性的要求出发,用有理数的“分割”来定义无理数,并把实数理论建立在严格的科学基础上,从而结束了无理数被认为“无理”的时代,也结束了持续2000多年的数学史上的第一次大危机。
参考资料:百度百科,无理数
证明根号3是无理数,使用反证法
如果√3是有理数,必有√3=p/q(p、q为互质的正整数)
两边平方:3=p^2/q^2
p^2=3q^2
显然p为3的倍数,设p=3k(k为正整数)
有9k^2=3q^2 即q^2=3k^2
于是q于是3的倍数,与p、q互质矛盾
∴假设不成立,√3是无理数
反证:假设根号3是有理数,则存在两个互质整数m和n使得根号3=m/n.两边平方并整理得m^2=3n^2,
于是m是3的倍数,令m=3q,
代入上式整理得:n^2=3q^2,
故n也是3的倍数,这与m,n互质矛盾。故根号3是无理数。证毕。
分析:
①有理数的概念:
“有限小数”和“无限循环小数”统称为有理数。
整数和分数也统称为有理数。
所有的分数都是有理数,分子除以分母,最终一定是循环的。
②无理数的概念:无限不循环小数,可引申为“开方开不尽的数”。
③反证法的要领是假设一个明显荒谬的结论成立,然后正确地证明原假设是错误的。
解:
假设(√3)是有理数,
∵ 1<3<4
∴(√1)<(√3)<(√4)
即:1<(√3)<2
∴(√3)不是整数。
∵整数和分数也统称为有理数,而(√3)不是整数
∴在假设“(√3)是有理数”的前提下,(√3)只能是一个分子分母不能约分的分数。
此时假设 (√3) = m/n(m、n均为正整数且互质,二者不能再约分,即二者除1外再无公因数)
两边平方,得:
m² / n² = 3
∴m² 是质数3的倍数
我们知道,如果两个数的乘积是3的倍数,那么这两个数当中至少有一个数必是3的倍数。
∴由“m² (m与m的乘积) 是质数3的倍数”得:正整数m是3的倍数。
此时不妨设 m = 3k(k为正整数)
把“m = 3k” 代入“m² / n² = 3” ,得:
(9k²) / n² = 3
∴3k² = n²
即:n² / k² = 3
对比“m² / n² = 3“ 同理可证
正整数n也是3的倍数
∴正整数m和n均为3的倍数
这与“m、n均为正整数且互质”相矛盾。
意即由原假设出发推出了一个与原假设相矛盾的结论,
∴原假设“(√3) = m/n(m、n均为正整数且互质,二者不能再约分,即二者除1外再无公因数)”是不成立的。
∴(√3) 不能是一个分子分母不能约分的分数
而已证(√3) 不是整数
∴(√3) 既 不是整数也不是分数,即(√3) 不是有理数。
∴(√3) 是无理数。
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