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

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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

(\binom{n}{0}) + (\binom{n}{1}) + (\binom{n}{2}) + . . . + (\binom{n}{n}) = 2^{n}

be P(n)

Assume this is true for some n, then

(\binom{n}{0}) + (\binom{n}{1}) + (\binom{n}{2}) + . . . + (\binom{n}{n}) = 2^{n}

2 \times {(\binom{n}{0}) + (\binom{n}{1}) + (\binom{n}{2}) + . . . + (\binom{n}{2})} = 2^{n + 1}

{(\binom{n}{0}) + (\binom{n}{n})} + {(\binom{n}{0}) + 2 (\binom{n}{1}) + 2 (\binom{n}{2}) + . . . + 2 (\binom{n}{n - 1}) + (\binom{n}{n})} = 2^{n + 1}

{(\binom{n}{0}) + (\binom{n}{n})} + {(\binom{n}{0}) + (\binom{n}{1})} + {(\binom{n}{1}) + (\binom{n}{2})} + {(\binom{n}{2}) + (\binom{n}{3})} + . . . + {(\binom{n}{n - 1}) + (\binom{n}{n})} = 2^{n + 1}

Now using the identities of this function:

(\binom{n}{a}) + (\binom{n}{a + 1}) = (\binom{n + 1}{a + 1})

(Note:If anyone find wikibooks ever mentioned this, include a link here!),we have:

{(\binom{n}{0}) + (\binom{n}{n})} + (\binom{n + 1}{1}) + (\binom{n + 1}{2}) + (\binom{n + 1}{3}) + . . . + (\binom{n + 1}{n}) = 2^{n + 1}

Since

(\binom{n}{0}) = (\binom{n}{n}) = 1

for all n,

(\binom{n + 1}{0}) + (\binom{n + 1}{n + 1}) + (\binom{n + 1}{1}) + (\binom{n + 1}{2}) + (\binom{n + 1}{3}) + . . . + (\binom{n + 1}{n}) = 2^{n + 1}

(\binom{n + 1}{0}) + (\binom{n + 1}{1}) + (\binom{n + 1}{2}) + (\binom{n + 1}{3}) + . . . + (\binom{n + 1}{n}) + (\binom{n + 1}{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} = (\binom{n}{0}) a^{n} + (\binom{n}{1}) a^{n - 1} b + \dots + (\binom{n}{n}) b^{n}

letting a = b = 1, we get

(1 + 1)^{n} = 2^{n} = (\binom{n}{0}) + (\binom{n}{1}) + \dots + (\binom{n}{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} (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.

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

Mathematical Proofs Problem Set

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