Separate it to 2 cases:
1) a is an integer;
2) a is not an integer.
Do you see what to do now?
Oh, and welcome to the forums
Hi,
I can't seem to wrap my head around this proof of rational numbers and integers.
The statement says:
For all rational numbers a, there is a rational number b, such that a + b is not an integer.
I say this statement is true.
I would suppose there is a rational number a, where a = p/q for some integers p and q.
But I'm not sure what to let b equal, in order for it to be a rational every time. I've tried letting b equal some manipulation of a, but nothing seem to work. For example, I tried letting b = 1/a or q/p. But it p was 0, then b is undefined and wouldn't work.
Can I get help on this please?
Thanks in advance!
Thanks for the quick reply!
So for the first case, where a is an integer, would it be okay if I wrote:
Case1: Suppose p divides q, and thus a is an integer.
Let b = x/y; for some integers x and y, and y cannot equal 0
(a) + (x/y) = (ay + x)/y; where (ay + x) is an integer
Therefore a + b is not an integer
Case2: Suppose p does not divide q, and thus a is not an integer
Let b = s/t; for some integers s and t, and t cannot equal 0
(a) + (s/t) = (at + s)/t; where (at + s) is an integer
Therefore a + b is not an integer
I really feel like I'm not being clear enough here, because (ay + x) could still be divisible by y and (at+s) could still be divisible by t, but did I fix that by saying for some x/y, and s/t?
And thank you again for the welcome!
Edit: After reading that through, I'm almost positive I did that wrong, and now I'm more lost than ever.
You're welcome.
You're not being clear enough - that's correct:
What you need to do is choose a number b, such that a+b is not an integer.
The purpose of separating the cases is that for each case, you can easily find a suitable b.
I'll help you with the easier one (well, they're both easy once you understand what you need to do ): If a is not an integer, you can choose b=0. Then a+b = a, and a is not integer, hence a+b is not an integer. Alternatively, for this case, you can choose any integer to be b, and that would work as well.
Now, what b must you choose, so that a+b is not an integer, when a is an integer?