We define even and odd as follows: an integer is even if for some integer , and is odd if is even. An immediate consequence of the definition, which we will feel free to use, is that an integer is odd if and only if for some integer .
- Prove than an integer cannot be both even and odd.
- Prove that every integer is either even or odd.
- Prove that the set of even integers is closed under sums and products. What can be said about sums and products of odd integers?
- Prove that if is even, must be even as well. Prove also that for integers and , if , then both and must be even.
- Prove that every rational number can be written as the ratio of integers and at least one of which is odd.
- Proof. Suppose otherwise, that there is some that is both even and odd. Then, there exist such that and (since is odd, is even, so ). Hence,
But , contradicting the fact that the integers are closed under addition (and subtraction)
- Proof. Let be an arbitrary integer. Then, is a real number, so by I.3.12, Exercise #4 we know there exists such that
If , then is even and we are done.
Otherwise, if , then we have
But this implies since is the only integer in the interval. Thus, is odd.
Hence, every integer is either even or odd
- Proof. Let be even integers. Then, there exist integers and such that , . Then,
where and are integers. Hence, and are both even
Next, we claim the sum of two odd integers is even, and the product of two odd integers is odd.
Proof. Let be odd integers. Then there exist integers and such that and . Then,
Hence, for some integer and for some integer . Therefore, is even and is odd, as claimed
- Proof. Let with even. Since and we know from part (b) that must be either even or odd and from part (a) that it cannot be both. By part (c) the product of two even integers even, while the product of two odd integers is odd; hence, we must have even (otherwise would be the product of two odd integers and would have to be odd).
Next, if and are integers with , then is even by definition (since it equals for some integer ). So, by the above we know is even, say . Then . But then we have,
and by the same reasoning that lead to even, we also have must be even. Thus, both and are even
- Proof. Let be a rational number with . If and are both even, then we have
with , and . Now, if and are both even, repeat the process. This will give a strictly decreasing sequence of positive integers, so the process must terminate by the well-ordering principle (see I.4.3, p. 34 in Apostol, Calculus, Volume I ). Thus, we must have some integers and , not both even with . This was the requested property
– Since any rational number has a greatest integer we can divide it in its greatest integer and its fraccional part s.t 0<(1/m)<=1 or -1<=(1/m)<0. If m is an integer (1/m) is a rational number.
– If n is an integer, n+(1/m) = (mn+1)/n and we have that (mn+1)/n is a rational number.
– assume n is even:
if m is odd (mn+1) is odd.
if m is even (mn+1) is odd.
How can a be even if a= bsqr2 ?
a = 0 and b = 0 is an example.
Hi… thanks for this… could the greatest lower bound theorem be used instead of the well-ordered principle? Restricting a/b>0 (extendable to a/b<0 realizying -1*a/b=a/b), the set S of numerators (or denominators) is non-empty and bounded below as they must be positive integers. The process will end whenever an odd value is obtained but it is guaranteed to end when it reaches Inf(S)=1… just an idea
I meant for part (e)
Is there any way of proving part(e) without using the well ordering principle and the principle of induction which are not explicitly given until the next section? BTW, thank you very much for sharing these excellent solutions. Great job!
*or the principle of induction
Prove the infimum of a set of integers is its min value, then create a set of numerators of the rational bounded below by 0. The min element and it’s corresponding denominator are assumed to be even, divide it and it’s corresponding denominator by 2 and you have a new integer numerator that should also be in the set because it’s a numerator of the rational, yet it shouldn’t be in the set because it’s below the minimum. Contradiction of the fact that both numerator and denominator are always even.