Help me on a math question QianYongQiang_section3_24

wetheliving

Cattt, Thanks a lot! In one of your previous reply post, you said, wu2li4shu4 can not be expressed as square root format. I guess you mean you3li3shu4 can not be expressed as square root? (Sorry, I do not have chinese input)

williamlee

如果n不是整数，那么存在p，q，使得  n=p/q,而且p，q的最大公约是是1（也就是说他们互素）
那么p^2，q^2也是互素的，因此n^2不可能是整数

thanks for the explanation!

catttt

补充一下，无理数是不能表示成根号形式的。虽然这个是可以证明的，但是只要记住结论就好了

[此贴子已经被作者于2004-9-13 23:37:44编辑过]

catttt

如果n不是整数，那么存在p，q，使得  n=p/q,而且p，q的最大公约是是1（也就是说他们互素）

那么p^2，q^2也是互素的，因此n^2不可能是整数

和下面一题相比，问题处在n是否有理数，如果n是有理数，那么他可以表示成分数形式，否则，存在无理数如根号2等，其平方是整数，而他本身不是。

wetheliving

If n=p/q, where p and q are nonzero integers, is n an integer?

1). n^2 is an integer

2). (2n+4)/2 is an integer

The answer is (D). I worked on this one for half an hour this morning and        could not figure out a way to prove that if n=p/q and n^2 is an integer then n must be an integer. Very frustrated :-)

Can Any NN        help me to understand the logic behind this question?

In the same section 3, question 15:

Is n an integer?

1). n^2 is an integer

2). squre root of n is an integer

Answer is B.

Comparing these two question, I think the trick must be because n can be expressed as one integer devided by another integer. Pls help me out!

[此贴子已经被作者于2004-9-13 23:18:07编辑过]