Prove that there is no such that .
Suppose otherwise, that there is some rational number
. Then, since
, we know there exist integers
not both even (I.3.12, Exercise #10 (e)
. Then, since
But, by I.3.12, Exercise #10 (d), we know implies both and are even, contradicting our choice of and not both even. Hence, there can be no such rational number
Let be given with . Prove that there exists an irrational number such that .
Note: To do this problem, I think we need to assume the existence of an irrational number. We will prove the existence of such a number (the ) in I.3.12, Exercise #12.
Proof. Since the rationals are dense in the reals I.3.12, Exercise #6, we know that for with there exist such that
Now, assume the existence of an irrational number, say (see note preceding the proof about this). Since we know and from the order axioms exactly one of or is positive ( is nonzero since ). Without loss of generality, let . Then, since , we know there exists an integer such that
Also, since , we have ; thus, .
Then, by I.3.12, Exercise #7 we have irrational and hence irrational.
Thus, letting , we have with irrational
Prove that the sum of two irrational numbers need not be irrational and the product of two irrational numbers need not be irrational.
be irrational. Then, by an argument we made in I.3.12, Exercise #7
, we know that
are both irrational. However,
Therefore, the sum and product of two irrationals need not be irrational
(Note: we could also just say is not irrational, but we have not yet established that is irrational. In fact, I’m not sure we’ve even established that an irrational number actually exists yet.)
Prove that if with , and , then
are all irrational, i.e., are all in .
, we know there are integers
. Now we consider each of the elements we wish to show are irrational.
- Suppose otherwise, that . Then, there exist such that
This contradicts our assumption that is irrational.
- Since is irrational, so is (by part (a), the sum of a rational and an irrational cannot be rational and since is rational, cannot have rational). But then, and by part (a) this sum must be irrational.
- Suppose otherwise, that . Then there exist such that . Further, since , we know exists.
Contradicting our assumption that is irrational.
- First, we since is irrational, we have , and thus exists. Further, since is rational, and is irrational, by (c) we must have irrational as well. Then by (c), since , we must have irrational.
- By (d) we know is irrational, and since is rational, by (c), we must have irrational.