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.