You are welcome to use any result of Theorem 3.1 and 3.2 in the book if you find them helpful.
Prove that \(1^3 + 2^3 + 3^3 + \cdots + n^3 = (1 + 2 + 3 + \cdots + n)^2\).
We prove this using induction. As a base case, we notice that \(1^3 = 1 = (1)^2\).
Suppose now that \(1^3 + 2^3 + \cdots + k^3 = (1 + 2 + \cdots k)^2\). Then:
\[\begin{align*} 1^3 + 2^3 + \cdots + k^3 + (k+1)^3 &= (1^3 + 2^3 + \cdots + k^3) + (k+1)^3 \\ &= (1 + 2 + \cdots k)^2 + (k+1)^3 \\ &= \left(\frac{k(k+1)}2\right)^2 + (k+1)^3 \\ &= (k+1)^2\left(\frac{k^2}4 + (k+1)\right) \\ &= (k+1)^2\left(\frac{k^2+4k+4}4\right) \\ &= (k+1)^2\left(\frac{(k+2)^2}4\right) \\ &= \left(\frac{(k+1)((k+1)+1)}2\right)^2 \\ &= (1 + 2 + \cdots (k+1))^2 \end{align*}\]So we are done by induction.
Guess a formula for the sum \(1 + 3 + \cdots + (2n - 1)\) (try evaluating for \(n = 1,2,3,4\) and look for patterns). Then prove that your formula is correct.
We claim that \(1 + 3 + \cdots + (2n - 1) = n^2\).
We prove this using induction. As a base case, we notice that \(1 = 1^2\).
Suppose now that \(1 + 3 + \cdots + (2k - 1) = k^2\). Then:
\[\begin{align*} 1 + 3 + \cdots + (2k - 1) + (2k + 1) &= (1 + 3 + \cdots + (2k - 1)) + (2k + 1) \\ &= k^2 + (2k - 1)\\ &= (k+1)^2 \end{align*}\]So we are done by induction.
For each subset of \(\mathbb R\) below, determine both the maximum and the minimum, if they exist. If either doesn’t exist, say so. You do not need to give a rigorous proof of your answer.
Maximum 5, minimum 1
Maximum 1/2, minimum -1
Maximum 1, minimum -1
Maximum DNE, minimum DNE
Maximum DNE, minimum 1/2
Don’t worry about writing out any formal proofs in this problem. Decide whether each of the following statements is true. If the statement is true, you don’t need to do anything more. If the statement is false, give a concrete example (that is, a counterexample) that shows the statement failing.
An irrational number is an element of \(\mathbb R \setminus \mathbb Q\). If \(r \ne 0\) is rational and \(\alpha\) is irrational, then \(r + \alpha\) is irrational.
\(\vert \vert a \vert - \vert b \vert \vert \le \vert a - b \vert\) for all \(a, b \in \mathbb R\).
A nonempty finite set always has a maximum.
True. If \(r + \alpha = p/q\) for some integers \(p, q\) then \(\alpha = p/q - r\), a rational; this is a contradiction.
True. Without loss of generality, assume that \(\vert a \vert \ge \vert b \vert\) so that \(\vert \vert a \vert - \vert b \vert \vert = \vert a \vert - \vert b \vert\). We have that \(\vert a \vert = \vert (a-b) + b \vert \le \vert a-b \vert + \vert b \vert\) by the Triangle inequality, which is equivalent to saying that \(\vert a \vert - \vert b \vert \le \vert a - b \vert\).
True. This is a fact about maximums; since the set is finite, a maximum can always be found in finite time (just look through all elements).
Let \(A\) be a nonempty set of real numbers which has a minimum. Let \(-A\) be the set \(\{ -x \;:\; x \in A \}\). Prove that
\[\min A = -\max(-A).\]Let \(m = \min A\). So for each \(a \in A, m \le a\). Then for any \(y=-a \in -A\), we have that \(m \le a\) and so \(-m \ge -a = y\). So \(-m = \max (-A)\) or equivalently that \(\min A = -\max(-A)\).