Math 290 Lecture #12
§5.4,5.5: More on Proofs, Part V
§5.4: Existence Proofs. A theorem about existence of some x ∈ S for which R(x) is
true has the logical form
∃x ∈ S, R(x).
We might be able to find or “construct” an actual value of x ∈ S for which R(x) is true.
Then again, we might be able to show that a value of x ∈ S exists for which R(x) is true,
but not in a constructible way.
For the latter, think about the theorem that every odd degree polynomial has a real root:
the proof relies on the intermediate value theorem of Calculus; it does not tell us how to
find the real root.
A proof of an existence theorem is called an existence proof.
√
Recall that we have seen that 2 = 21/2 is irrational.
We have two rational numbers a = 2 and b = 1/2 for which ab is irrational.
Thus we an existence proof that there are rational numbers a and b for which ab is
irrational..
What about other possibilities, such as there exists irrational numbers a and b such that
ab is rational?
Result 5.21. There exist irrational numbers a and b such that ab is rational.
√
2.
√
Until we have more irrational numbers available, we will have to use 2.
√ √2
Proof. Consider 2 .
For a proof strategy, we know of only one irrational number, namely
As we are dealing with rational and irrational numbers here, there are two cases.
√ √2
Case 1 2 is rational.
Then we have the desired result.
√ √2
Case 2 2 is irrational.
b
Then raising the irrational number a =
which is rational.
√ √2
2
√2
√
√
2
2
to the irrational number
√
2 gives
√ √√
√ 2
= ( 2) 2 2 = 2 = 2
Now you feel that this prove is lacking in that it does not specify a pair of irrational
numbers a and b for which ab is rational.
√ √2
It is know that 2 is irrational.
We now add something to the existence proof, namely the idea of uniqueness: there is
only one x ∈ S for which R(x) is true.
Not all quantified statements ∃x ∈ S, R(x) have exactly one x ∈ S for which R(x) is
true.
This is the case with the intermediate value theorem, and the existence of a real root of
an odd-degree polynomial.
But there are situations where uniqueness can be obtained by proof.
There are two approaches to getting uniqueness:
(a) Assume a, b ∈ S such that R(a) and R(b) are true, and show that a = b by a direct
proof;
(b) Assume a, b ∈ S are distinct with R(a) and R(b) are true, and show that a = b or
some other contradiction.
Results 5.22, 5.23. The equation x5 + 2x − 5 = 0 has a unique real number solution
between x = 1 and x = 2.
Proof. Let f (x) = x5 + 2x − 5.
This polynomial function is continuous, and f (1) = −2 and f (2) = 31.
By the Intermediate Value Theorem there is c ∈ (1, 2) such that f (c) = 0.
To show uniqueness of this real root, suppose that a and b are two real roots of f (x) = 0
with 1 < a < b < 2.
The inequality 0 < a < b implies a5 < b5 , 2a < 2b, so that
a5 + 2a − 5 < b5 + 2b − 5.
However, since f (a) = 0 and f (b) = 0, we obtain
0 = a5 + 2a − 5 < b5 + 2b − 5 = 0,
which is a contradiction.
Result 5.24. For an irrational number r, let
S = {sr + t : s, t ∈ Q}.
For every x ∈ S there exist unique rational numbers a and b such that x = ar + b.
Proof. The existence is by the definition of the set S.
To show uniqueness, suppose for x ∈ S we have x = ar + b and x = cr + d where
a, b, c, d ∈ Q and a 6= c.
We do not need to assume that b 6= d because a 6= c is enough to guarantee that ar + b
and cr + d are different.
We take ar + b = x = cr + d and write it as
(a − c)r = d − b.
With a 6= c, we can solve for r getting
r=
d−b
.
a−c
Since a, b, c, d ∈ Q, the ratio r = (d − b)/(a − c) ∈ Q, a contradiction to r being rational.
Thus a = c from which it follows that ar = cr, so that b = d.
§5.5: Disproving Existence Statements. To prove ∃x ∈ S, R(x) is not true we
prove that its negation, ∀x ∈ S, ∼ R(x) is true.
Example 5.26. Disprove that there exists an odd integer n such that n2 + 2n + 3 is
odd.
We show that for all odd integers n that n2 + 2n + 3 is even.
For n odd we have n = 2k + 1 for some k ∈ Z.
Then
n2 + 2n + 3 = (2k + 1)2 + 2(2k + 1) + 3
= 4k 2 + 4k + 1 + 4k + 2 + 3
= 4k 2 + 8k + 6
= 2(2k 2 + 4k + 3).
Since 2k 2 + 4k + 2 ∈ Z, we have that n2 + 2n + 3 is even.
Example. Disprove that there exists an odd integer n such that 8 - (n2 + (n + 6)2 + 6).
To disprove this, we prove that for all odd integers n we have 8 | (n2 + (n + 6)2 + 6).
For n odd we have n = 2k + 1 for some k ∈ Z, and so
n2 + (n + 6)2 + 6 = (2k + 1)2 + (2k + 1 + 6)2 + 6
= 4k 2 + 4k + 1 + 4k 2 + 28k + 49 + 6
= 8k 2 + 32k + 56
= 8(k 2 + 4k + 7).
Since k 2 + 4k + 7 ∈ Z, we have that 8 | (n2 + (n + 6)2 + 6).