Mathski — Proof

Proof

IB Mathematics AA · HL · Shadow Worksheet

Practice

Name

Question 1

Prove that the product of two odd numbers is always odd.

Let two odd numbers be

\(2m+1\) and \(2n+1\) for integers \(m,n\)

Product

\((2m+1)(2n+1)=4mn+2m+2n+1=2(2mn+m+n)+1\)

Conclusion

This is of the form \(2k+1\) where \(k=2mn+m+n\in\mathbb{Z}\). Hence the product is odd. ■

Question 2

Prove that if \(n\) is even then \(n^3+n\) is divisible by 4.

Let \(n=2k\)

\(n^3+n=8k^3+2k=2k(4k^2+1)\)

Since \(n=2k\) is even

\(n^3+n=2k(4k^2+1)\). For this to be divisible by 4, we need \(2k\) divisible by 4 OR \(4k^2+1\) even — but \(4k^2+1\) is always odd.

Re-approach

\(n^3+n=n(n^2+1)\). With \(n=2k\): \(2k(4k^2+1)\). Note \(4k^2+1\) is odd; so we need \(2k\) to supply the factor of 4. When \(k\) is any integer, \(n(n^2+1)=2k(4k^2+1)\) is divisible by 2. We need to show divisible by 4: take \(n=2k\), then \(n^2+1=4k^2+1\equiv 1\pmod{4}\), and \(n=2k\equiv 2\pmod 4\) when \(k\) is odd, giving product \(\equiv 2\pmod 4\) — not always divisible by 4. Counterexample: \(n=2\): \(8+2=10\), not divisible by 4.

Conclusion

Statement is false. Counterexample: \(n=2\) gives \(n^3+n=10\), which is not divisible by 4.

Question 3

Prove by contradiction that \(\sqrt{3}\) is irrational.

Assume

\(\sqrt{3}=\dfrac{p}{q}\) in lowest terms (\(\gcd(p,q)=1\))

Then

\(3=\dfrac{p^2}{q^2}\Rightarrow p^2=3q^2\), so \(3|p^2\Rightarrow 3|p\). Let \(p=3k\).

Substitute

\(9k^2=3q^2\Rightarrow q^2=3k^2\Rightarrow 3|q\). But then \(3|p\) and \(3|q\), contradicting \(\gcd(p,q)=1\).

Conclusion

\(\sqrt{3}\) is irrational. ■

Question 4

Prove by induction that \(\displaystyle\sum_{r=1}^{n} r^2 = \dfrac{n(n+1)(2n+1)}{6}\).

Base case \(n=1\)

LHS: \(1\); RHS: \( frac{1\cdot2\cdot3}{6}=1\). ✓

Inductive step

Assume true for \(n=k\): \(\sum_{r=1}^k r^2= frac{k(k+1)(2k+1)}{6}\).

For \(n=k+1\)

\(\sum_{r=1}^{k+1}r^2= frac{k(k+1)(2k+1)}{6}+(k+1)^2= frac{(k+1)[k(2k+1)+6(k+1)]}{6}= frac{(k+1)(2k^2+7k+6)}{6}= frac{(k+1)(k+2)(2k+3)}{6}\) ✓

Conclusion

By the principle of mathematical induction, the formula holds for all \(n\in\mathbb{Z}^+\). ■

Question 5

Prove by induction that \(7^n - 1\) is divisible by 6 for all \(n\in\mathbb{Z}^+\).

Base case \(n=1\)

\(7-1=6\). Divisible by 6. ✓

Inductive step

Assume \(7^k-1=6m\) for some integer \(m\), so \(7^k=6m+1\).

For \(n=k+1\)

\(7^{k+1}-1=7\cdot7^k-1=7(6m+1)-1=42m+6=6(7m+1)\). Divisible by 6. ✓

Conclusion

By induction, \(7^n-1\) is divisible by 6 for all \(n\in\mathbb{Z}^+\). ■

Generated by Mathski · mathski.io · IB Mathematics AA Shadow Worksheets