# Lecture 009

## Biconditional Claims

Claim: for any a and b, a and b have the same parity (they are all even or all odd) iff there exists $m,n \in \mathbb{Z}(a=m+n \land b=m-n)$

$(\forall a,b \in \mathbb{Z})((\exists k, l \in \mathbb{Z})((a=2k \land b=2l)\lor (a=2k+1 \land b=2l+1)) \equiv (\exists m,n \in \mathbb{Z})(a-m+n \land b=m-n))$

Proof:

• (\rightarrow)

• case: a and b are both even -> find m and n in terms of k and l by algebra manipulation.
• case: a and b are odd -> same as above
• (\leftarrow)

• AFSOC $(\exists m, n \in \mathbb{Z})(a=m+n \land b=m-n)$ but a, b have opposite parities. WLOG Assuming one is odd, the other is even, manipulate algebraic equations into $2(k+l)+1-2m$ to show $k+l-m=-\frac{1}{2}$ and that is a contradiction since $-\frac{1}{2}$ is not an integer.

Claim: for all integers m and n $m^2(n^2-2n)$ is odd iff m and n are both odd. Proof: using definition only. easy, same as above

Table of Content