Answer
See solution
Work Step by Step
### Part (a): Prove that for all integers \(a\), if \(a^3\) is even then \(a\) is even.
**Proof (by contrapositive):**
The contrapositive of “if \(a^3\) is even then \(a\) is even” is “if \(a\) is odd then \(a^3\) is odd.” We prove this contrapositive statement.
1. Suppose \(a\) is odd. Then by definition there exists an integer \(k\) such that:
\[
a = 2k + 1.
\]
2. Then,
\[
a^3 = (2k + 1)^3.
\]
3. Expanding using the binomial theorem:
\[
(2k+1)^3 = 8k^3 + 12k^2 + 6k + 1 = 2(4k^3 + 6k^2 + 3k) + 1.
\]
4. The expression \(2(4k^3 + 6k^2 + 3k)\) is even, and adding 1 yields an odd number. Hence, \(a^3\) is odd.
Since we have shown that if \(a\) is odd then \(a^3\) is odd, the contrapositive holds. Therefore, the original statement is true: if \(a^3\) is even then \(a\) must be even.
---
### Part (b): Prove that \(2^{\frac{1}{3}}\) is irrational.
**Proof by contradiction:**
1. Suppose, for the sake of contradiction, that \(2^{\frac{1}{3}}\) is rational. Then it can be written in lowest terms as:
\[
2^{\frac{1}{3}} = \frac{p}{q},
\]
where \(p\) and \(q\) are integers, \(q \neq 0\), and the fraction \(\frac{p}{q}\) is in its simplest form (i.e., \(p\) and \(q\) have no common factors).
2. Cube both sides of the equation:
\[
2 = \left(\frac{p}{q}\right)^3 = \frac{p^3}{q^3}.
\]
Multiplying both sides by \(q^3\) gives:
\[
p^3 = 2q^3.
\]
3. This equation shows that \(p^3\) is even (since it equals \(2q^3\)). By the result of part (a), if \(p^3\) is even then \(p\) is even. Therefore, we can write:
\[
p = 2r,
\]
for some integer \(r\).
4. Substitute \(p = 2r\) into the equation \(p^3 = 2q^3\):
\[
(2r)^3 = 2q^3 \quad \Longrightarrow \quad 8r^3 = 2q^3.
\]
Divide both sides by 2:
\[
4r^3 = q^3.
\]
This shows that \(q^3\) is even, and by the same argument as in part (a), \(q\) must be even.
5. Now, both \(p\) and \(q\) are even, meaning they have a common factor of 2. This contradicts the assumption that \(\frac{p}{q}\) was in lowest terms.
Since our assumption that \(2^{\frac{1}{3}}\) is rational leads to a contradiction, we conclude that \(2^{\frac{1}{3}}\) is irrational.
