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

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Mathematical Proofs Problem Set[edit]

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.