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

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Mathematical Proofs Problem Set[edit]


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.