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

Mathematical Proofs Problem Set



For all
Therefore   ,   ,  ...  
When a>b and c>d, a+c>b+d ( See also Replace it if you find a better one).
Therefore we have:


Let us call the proposition
  be P(n)
Assume this is true for some n, then
Now using the identities of this function: (Note:If anyone find wikibooks ever mentioned this, include a link here!),we have:
Since   for all n,
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


letting a = b = 1, we get


as required.


Let   be a polynomial with x as the variable, y and n as constants.
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
  is an integer for all integer value of x,y and n when n is odd.