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Math 310: Introduction to Mathematical Reasoning Summer 2006 Homework 4 Solutions
(From An Introduction to Mathematical Reasoning, by Peter J. Eccles.)
- Prove by induction on n that, for all positive integers n, n^3 − n is divisible by 3.
Proof. Consider the base case n = 1. 13 − 1 = 0 = 3 · 0, so 1^3 − 1 is divisible by 3. Now suppose k^3 − k is divisible by 3 for some k ≥ 1. Then k^3 − k = 3q for some positive integer q. (k+1)^3 −(k+1) = k^3 +3k^2 +3k+1−k−1 = (k^3 −k)+(3k^2 +3k) = 3q+3(k^2 +k) = 3(q+k^2 +k), so (k + 1)^3 − (k + 1) is divisible by 3. By induction, n^3 − n is divisible by 3 for all positive n.
- Prove by induction on m that m^3 ≤ 2 m^ for m ≥ 10.
Proof. Certainly, for the base case, 10^3 = 1000 ≤ 1024 = 2^10. Now suppose that for some k ≥ 10, k^3 ≤ 2 k. To prove that (k +1)^3 ≤ 2 k+1^ = 2· 2 k, it suffices to prove that (k +1)^3 ≤ 2 k^3 , or, expanding (k + 1)^3 , that, k^3 + 3k^2 + 3k + 1 ≤ 2 k^3 , or equivalently that 3k^2 + 3k + 1 ≤ k^3. We will have established this if we can show that each term in the sum on the left is less than or equal to 13 k^3 individually. But 3k^2 ≤ 13 10 k^2 ≤ 13 k^3 , and 3k ≤ 13 10 k ≤ 13 k^2 ≤ 13 k^3 , and 1 ≤ 13 10 ≤ 13 k ≤ 13 k^3 , so indeed 3k^2 + 3k + 1 ≤ k^3 , as desired. By induction, the proposition is true for all m ≥ 10.
- Prove by induction on n that, for all positive integers n, n ≥ 1.
Proof. It is certainly true that 1 ≥ 1, establishing the base case. Now suppose that k ≥ 1. We will use the fact that 1 ≥ 0, as we have previously proven. k + 1 ≥ k + 0 = k and k ≥ 1, so by transitivity k + 1 ≥ 1. By induction, n ≥ 1 for all positive integers n.
- Prove by induction on n that, for any real numbers x and y and for non-negative integers m and n:
(a) xm+n^ = xmxn;
Proof. Consider the base case n = 0; by definition, x^0 = 1, so for any m, xm+0^ = xm^ = xmx^0 , as desired. Now suppose that for some k ≥ 0, xm+k^ = xmxk^ for all non-negative integers m. Then for any such m, xm+(k+1)^ = x(m+k)+1^ = xm+kx = xmxkx = xmxk+ where we have used the inductive definition of exponentiation repeatedly. By induction, for any nonnegative integers m and n, xm+n^ = xmxn.
(b) (xm)n^ = xmn.
Proof. Consider the base case n = 0; by definition, for any non-negative integer m, (xm)n^ = 1 = x^0 = xm·^0. Now suppose that for some k ≥ 0, for all non-negative integers m, (xm)k^ = xmk. Then (xm)k+1^ = (xm)kxm^ = xmkxm^ = xmk+m^ = xm(k+1)^ by the previous exercise. By induction, (xm)n^ = xmn^ for all non-negative integers m and n.
[These statements are known as the laws of exponents or the laws of indices.]
(You may assume xnyn^ = (xy)n.)
(From Mathematical Thinking, by John P. D’Angelo and Douglas B. West.)
- For n ∈ N, prove that |
∑n i=1 ai| ≤^
∑n i=1 |ai|. (You may assume that, for any a, b ∈ R, |a + b| ≤ |a| + |b|. This is known as the Triangle Inequality.)
Proof. The base case n = 1 is trivial: |
i=1 ai|^ =^ |a^1 |^ =^
i=1 |ai|.^ Now suppose that |
∑k i=1 ai| ≤^
∑k i=1 |ai|^ for some^ k^ ≥^ 1.^ Then^ |^
∑k+ i=1 ai|^ =^ |^
∑k i=1 ai^ +^ ak+1| ≤ |^
∑k i=1 ai|^ + |ak+1| ≤
∑k i=1 |ai|^ +^ |ak+1|^ =^
∑k+ i=1 |ai|. By induction, the proposition holds for all positive integers n.
