High School Mathematics Extensions/Mathematical Proofs/Problem Set/Solutions

Mathematical Proofs Problem Set
1.
 * For all

\begin{matrix} a & > & 0\\ n+a & > & n\\ n & > & n-a\\ \sqrt{n} & > & \sqrt{n-a}\\ 1 & > & \frac{\sqrt{n-a}}{\sqrt{n}}\\ \frac{1}{\sqrt{n-a}} & > & \frac{1}{\sqrt{n}} \end{matrix} $$
 * Therefore $$\frac{1}{\sqrt{1}}$$, $$\frac{1}{\sqrt{2}}$$ , $$\frac{1}{\sqrt{3}}$$... $$ > \frac{1}{\sqrt{n}}$$
 * When a>b and c>d, a+c>b+d ( See also Replace it if you find a better one).
 * Therefore we have:
 * $$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}......+\frac{1}{\sqrt{n}}>n\times\frac{1}{\sqrt{n}}$$
 * $$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}......+\frac{1}{\sqrt{n}}>\frac{n}{\sqrt{n}}\times\frac{\sqrt{n}}{\sqrt{n}}$$
 * $$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}......+\frac{1}{\sqrt{n}}>\frac{n\sqrt{n}}{n}$$
 * $$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}......+\frac{1}{\sqrt{n}}>\sqrt{n}$$

3.
 * Let us call the proposition
 * $${n \choose 0} + {n \choose 1} + {n \choose 2} + ... + {n \choose n} = 2^n$$ be P(n)
 * Assume this is true for some n, then
 * $${n \choose 0} + {n \choose 1} + {n \choose 2} + ... + {n \choose n} = 2^n$$
 * $$2\times \left \{ {n \choose 0} + {n \choose 1} + {n \choose 2} + ... + {n \choose 2} \right \} = 2^{n+1}$$
 * $$\left \{ {n \choose 0} + {n \choose n} \right \} + \left \{ {n \choose 0} + 2{n \choose 1} + 2{n \choose 2} + ... + 2{n \choose n-1} + {n \choose n} \right \} = 2^{n+1}$$
 * $$\left \{ {n \choose 0} + {n \choose n} \right \} + \left \{ {n \choose 0} + {n \choose 1} \right \} + \left \{ {n \choose 1} + {n \choose 2} \right \} + \left \{ {n \choose 2} + {n \choose 3} \right \} + ... + \left \{ {n \choose n-1} + {n \choose n} \right \} = 2^{n+1}$$
 * Now using the identities of this function:$${n \choose a} + {n \choose a+1} = {n+1 \choose a+1}$$(Note:If anyone find wikibooks ever mentioned this, include a link here!),we have:
 * $$\left \{ {n \choose 0} + {n \choose n} \right \} + {n+1 \choose 1} + {n+1 \choose 2} + {n+1 \choose 3} + ... + {n+1 \choose n} = 2^{n+1}$$
 * Since $${n \choose 0} = {n \choose n} = 1$$ for all n,
 * $${n+1 \choose 0} + {n+1 \choose n+1} + {n+1 \choose 1} + {n+1 \choose 2} + {n+1 \choose 3} + ... + {n+1 \choose n} = 2^{n+1}$$
 * $${n+1 \choose 0} + {n+1 \choose 1} + {n+1 \choose 2} + {n+1 \choose 3} + ... + {n+1 \choose n} + {n+1 \choose n+1} = 2^{n+1}$$
 * Therefore P(n) implies P(n+1), and by simple substitution P(0) is true.
 * Therefore by the principal of mathematical induction, P(n) is true for all n.

Alternate solution Notice that
 * $$(a + b)^n = {n \choose 0} a^n + {n \choose 1} a^{n-1}b + \cdots + {n \choose n} b^n$$

letting a = b = 1, we get
 * $$(1 + 1)^n = 2^n = {n\choose 0} + {n \choose 1} + \cdots + {n\choose n}$$

as required.

5.
 * Let $$P(x)=x^n + y^n\,$$ be a polynomial with x as the variable, y and n as constants.

\begin{matrix} P(-y) & = & (-y)^n + y^n\\ \ & = & -y^n + y^n(\mbox{When n is an odd integer})\\ \ & = & 0 \end{matrix} $$
 * Therefore by factor theorem(link here please), (x-(-y))=(x+y) is a factor of P(x).
 * Since the other factor, which is also a polynomial, has integer value for all integer x,y and n (I've skipped the part about making sure all coeifficients are of integer value for this moment), it's now obvious that
 * $$\frac{x^n+y^n}{x+y}$$ is an integer for all integer value of x,y and n when n is odd.