P1.1bProof by contradiction

Answer: If \(n\) were odd, \(n = 2k + 1\) and \(n^2 = 2(2k^2 + 2k) + 1\) would be odd.

• B1 for assuming that \(n^2\) is even and \(n\) is odd
• M1 for writing \(n = 2k + 1\) and squaring: \(n^2 = 4k^2 + 4k + 1\)
• A1* for \(n^2 = 2(2k^2 + 2k) + 1\), which is odd, contradicting \(n^2\) even; so \(n\) is even

Worked solution: Assume \(n^2\) is even and \(n\) is odd, so \(n = 2k + 1\) for some integer \(k\).
Then \(n^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1\), which is odd.
This contradicts \(n^2\) being even, so \(n\) must be even.

Answer: \(\sqrt{2} = \frac{a}{b}\) in lowest terms gives \(a^2 = 2b^2\), so \(a\) is even, then \(b\) is even: contradiction.

• B1 for assuming \(\sqrt{2} = \frac{a}{b}\), where \(a\) and \(b\) are integers with no common factors, \(b \ne 0\)
• M1 for squaring and rearranging: \(a^2 = 2b^2\)
• A1 for deducing \(a^2\) is even, so \(a\) is even, and writing \(a = 2c\)
• M1 for substituting: \(4c^2 = 2b^2\), so \(b^2 = 2c^2\)
• A1* for \(b^2\) even so \(b\) even; \(a\) and \(b\) have a common factor 2, contradicting the assumption, so \(\sqrt{2}\) is irrational

Worked solution: Assume \(\sqrt{2}\) is rational: \(\sqrt{2} = \frac{a}{b}\) with \(a\), \(b\) integers having no common factors.
Then \(2 = \frac{a^2}{b^2}\), so \(a^2 = 2b^2\). Hence \(a^2\) is even, so \(a\) is even: \(a = 2c\).
Then \(4c^2 = 2b^2\), so \(b^2 = 2c^2\). Hence \(b^2\) is even and so \(b\) is even.
So \(a\) and \(b\) share the factor 2, contradicting the assumption. Therefore \(\sqrt{2}\) is irrational.

(a) Answer: If \(n = 3k \pm 1\) then \(n^2 = 3(3k^2 \pm 2k) + 1\), which is not a multiple of 3.

• M1 for considering \(n\) not a multiple of 3: \(n = 3k + 1\) or \(n = 3k + 2\) (or \(3k - 1\))
• A1 for \((3k + 1)^2 = 3(3k^2 + 2k) + 1\) and \((3k + 2)^2 = 3(3k^2 + 4k + 1) + 1\)
• A1* for neither is a multiple of 3, so if \(n^2\) is a multiple of 3 then \(n\) must be a multiple of 3

Worked solution: Suppose \(n\) is not a multiple of 3. Then \(n = 3k + 1\) or \(n = 3k + 2\) for some integer \(k\).
\((3k + 1)^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1\) and \((3k + 2)^2 = 9k^2 + 12k + 4 = 3(3k^2 + 4k + 1) + 1\).
Neither is a multiple of 3. So if \(n^2\) is a multiple of 3, \(n\) must be a multiple of 3.

(b) Answer: \(\sqrt{3} = \frac{a}{b}\) in lowest terms gives \(a^2 = 3b^2\), so 3 divides \(a\), then 3 divides \(b\): contradiction.

• B1 for assuming \(\sqrt{3} = \frac{a}{b}\), where \(a\) and \(b\) are integers with no common factors, \(b \ne 0\)
• M1 for \(a^2 = 3b^2\), so \(a^2\) is a multiple of 3 and, by part (a), \(a\) is a multiple of 3
• A1 for \(a = 3c\)
• M1 for \(9c^2 = 3b^2\), so \(b^2 = 3c^2\) and, by part (a), \(b\) is a multiple of 3
• A1* for \(a\) and \(b\) have a common factor 3, a contradiction, so \(\sqrt{3}\) is irrational

Worked solution: Assume \(\sqrt{3} = \frac{a}{b}\) with \(a\), \(b\) integers with no common factors.
Then \(a^2 = 3b^2\), so \(a^2\) is a multiple of 3 and by part (a) \(a\) is a multiple of 3: \(a = 3c\).
Then \(9c^2 = 3b^2\), so \(b^2 = 3c^2\) is a multiple of 3, and \(b\) is a multiple of 3.
So 3 is a common factor of \(a\) and \(b\): a contradiction. Hence \(\sqrt{3}\) is irrational.