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Math 310/ 3–11–
Solutions to Problem Set 9
[page 124]
Exercise 5.8. Prove that there is no smallest positive irrational number.
Suppose that x is the smallest positive irrational number. Consider
x.
Since x > 0, I have
x > 0, so
x is positive.
Since 1 >
, I have x >
x.
Finally, I’ll show that
x is irrational. Suppose on the contrary that
x is rational. Then I can write 1 2
x =
p q
, where p, q ∈ Z. Hence,
x =
2 p q
But 2p, q ∈ Z, so this shows that x =
2 p q
is rational, contradicting the assumption that x is irrational.
Therefore,
x is irrational.
All together,
x is a positive irrational which is smaller than x, but x was assumed to be the smallest
positive irrational number. This contradiction shows that there is no such x — that is, there is no smallest positive irrational number.
Exercise 5.12. Prove that 1000 cannot be written as the sum of three integers, an even number of which are even.
Suppose on the contrary that 1000 can be written as the sum of three integers, an even number of which are even. Say 1000 = a + b + c. There are two cases: Either the number of even summands is 0 or it is 2. Suppose there are 0 even summands. Then all three summands are odd, so
a = 2m + 1, b = 2n + 1, c = 2p + 1.
Here m, n, p ∈ Z. Then 1000 = a + b + c = (2m + 1) + (2n + 1) + (2p + 1) = 2(m + n + p + 1) + 1 But 2(m + n + p + 1) + 1 is odd, whereas 1000 is even. This is a contradiction. Suppose there are 2 even summands. Then there is one odd summand. There’s no harm in assuming that a and b are even and c is odd, since the names of the three summands are arbitrary. So
a = 2m, b = 2n, c = 2p + 1.
Here m, n, p ∈ Z. Then 1000 = a + b + c = 2m + 2n + (2p + 1) = 2(m + n + p) + 1
But 2(m + n + p) + 1 is odd, whereas 1000 is even. This is a contradiction. Since every case leads to a contradiction, it follows that 1000 cannot be written as the sum of three integers, an even number of which are even.
Exercise 5.20. Prove that if x and y are positive real numbers, then
x + y 6 =
x +
y.
Suppose that x and y are positive real numbers and √ x + y =
x +
y.
Then (
x + y)^2 = (
x +
y)^2 x + y = x + 2
x
y + y 0 = 2
x
y
Since x > 0 and y > 0, it follows that
x > 0 and
y > 0. Hence, 2
x
y > 0, which contradicts the last equation above. Hence, if x and y are positive real numbers, then
x + y 6 =
x +
y.
[page 83]
Exercise 3.20. Prove that if n ∈ Z, then n^3 − n is even.
Suppose n is even. Then n = 2m for some m ∈ Z. Hence,
n^3 − n = (2m)^3 − 2 m = 8m^3 − 2 m = 2(4m^3 − m)
Therefore, n^3 − n is even if n is even. Suppose n is odd. Then n = 2m + 1 for some m ∈ Z. Hence,
n^3 − n = (2m + 1)^3 − (2m + 1) = (8m^3 + 12m^2 + 6m + 1) − (2m + 1) = 8m^3 + 12m^2 + 4m = 2(4m^3 + 6m^2 + 2m)
Therefore, n^3 − n is even if n is odd. Hence, if n ∈ Z, then n^3 − n is even.
- Premises:
∼ Q → P
∼ Q ∨ R
R → P
Prove: P.
- ∼ Q → P Premise
- ∼ Q ∨ R Premise
- R → P Premise
- ∼ P Premise for proof by contradiction
- Q Modus tollens (1,4)
Case 1. x ≤ 1.
In this case, x − 1 ≤ 0, so |x − 1 | = −(x − 1). Also, x ≤ 1 certainly implies x ≤ 5, so x − 5 ≤ 0, and |x − 5 | = −(x − 5). Therefore,
|x − 1 | − |x − 5 | = −(x − 1) − [−(x − 5)] = − 4.
Thus, in this case, − 4 ≤ |x − 1 | − |x − 5 | ≤ 4 is true.
Case 2. 1 < x ≤ 5.
In this case, x − 1 > 0, so |x − 1 | = x − 1. Also, x ≤ 5, so x − 5 ≤ 0, and |x − 5 | = −(x − 5). Therefore,
|x − 1 | − |x − 5 | = (x − 1) − [−(x − 5)] = 2x − 6.
Now 1 < x ≤ 5 gives 2 < 2 x ≤ 10, so − 4 < 2 x − 6 ≤ 4. Thus, in this case, − 4 ≤ |x − 1 | − |x − 5 | ≤ 4 is true.
Case 3. x > 5.
In this case, x − 5 > 0, so |x − 5 | = x − 5. Also, x > 5 certainly implies x > 1, so x − 1 > 0, and |x − 1 | = x − 1. Therefore, |x − 1 | − |x − 5 | = (x − 1) − (x − 5) = 4. Thus, in this case, − 4 ≤ |x − 1 | − |x − 5 | ≤ 4 is true.
Since the inequality holds in all cases, it is true.
[page 124]
Exercise 5.14. Prove that when an irrational number is divided by a (nonzero) rational number, the resulting number is irrational.
Let x be irrational, and let
p q
be rational with p, q ∈ Z. I want to show that
x p q
is irrational.
Suppose on the contrary that
x p q
is rational. Then
x p q
m n , where m, n ∈ Z. Hence,
p q
x p q
p q
m n
x = pm qn
Since pm, qn ∈ Z, it follows that x =
pm qn is rational. This contradicts the assumption that x is irrational.
Therefore, x p q
is irrational.
[page 83]
Exercise 3.24. Let a, b ∈ Z. Prove that if a + b and ab are of the same parity, then a and b are even.
Suppose that a + b and ab are of the same parity. Suppose on the contrary that a and b are not both even. There are three cases.
Suppose a is even and b is odd. Then a = 2m and b = 2n + 1, where m, n ∈ Z. Then
a + b = 2m + (2n + 1) = 2(m + n) + 1
Hence, a + b is odd. Also, ab = (2m)(2n + 1) = 4mn + 2m = 2(2mn + m) Hence, ab is even. This contradicts the assumption that a + b and ab have the same parity. Suppose a is odd and b is even. Then a = 2m + 1 and b = 2n, where m, n ∈ Z. Then
a + b = (2m + 1) + 2n = 2(m + n) + 1
Hence, a + b is odd. Also, ab = (2m + 1)(2n) = 4mn + 2n = 2(2mn + n) Hence, ab is even. This contradicts the assumption that a + b and ab have the same parity. Suppose a is odd and b is odd. Then a = 2m + 1 and b = 2n + 1, where m, n ∈ Z. Then
a + b = (2m + 1) + (2n + 1) = 2(m + n + 1)
Hence, a + b is even. Also, ab = (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1
Hence, ab is odd. This contradicts the assumption that a + b and ab have the same parity. Since each of these three cases gives a contradiction, it follows that a and b are both even.
I prefer the errors of enthusiasm to the indifference of wisdom. - Anatole France
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