1. Prove that if $$a+b \ge 0$$ and $$ab \ge 0$$, then $$a \ge 0$$ and $$b \ge 0$$.
2. Prove that for all integers $$a$$ and $$b$$, $$a^2 - 4ab \ne 2$$.
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### Problem 1
If $$a+b \ge 0$$ and $$ab \ge 0$$, then $$a \ge 0$$ and $$b \ge 0$$.
#### Scratchwork
Direct proof? Hmm, not sure if it works.\
Contrapositive? Maybe:
> If $$a < 0$$ or $$b < 0$$, then $$a + b < 0$$ or $$ab < 0$$.
"$$a < 0$$ or $$b < 0$$" sounds like splitting the proof into *cases*. Plausible?\
Examples: $$a = -1$$, $$b = -1$$. Then $$ab = 1 > 0$$ but $$a + b = -2 < 0$$. More generally, if $$a < 0$$ and $$b \le 0$$ then $$a + b \le a + 0 = a < 0$$. What if $$b > 0$$? Then $$ab < 0$$ (multiplying positive and negative number).
#### Proof.
We will prove the contrapositive,
> If $$a < 0$$ or $$b < 0$$, then $$a + b < 0$$ or $$ab < 0$$.
There are two cases: $$a < 0$$ and $$b < 0$$.
Case $$a < 0$$:\
There are two subcases: $$b \le 0$$ and $$b > 0$$. If $$a < 0$$ and $$b \le 0$$ then $$a + b \le a + 0 = a < 0$$. If $$a < 0$$ and $$b > 0$$ then $$ab < 0$$.
Case $$b < 0$$:\
There are two subcases: $$a \le 0$$ and $$a > 0$$. If $$b < 0$$ and $$a \le 0$$ then $$a + b \le 0 + b = b < 0$$. If $$b < 0$$ and $$a > 0$$ then $$ab < 0$$.
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### Problem 2
For all integers $$a$$ and $$b$$, $$a^2 - 4ab \ne 2$$.
(Two different proofs)
#### Proof 1.
For the sake of contradiction, assume $$a^2 - 4ab = 2$$.\
Rearranging, we get $$a^2 = 2 - 4ab = 2(1-2ab)$$. The definition of even numbers now says that $$a^2$$ is even. Proposition 7 of the Proof notes says that if $$a^2$$ is even, then $$a$$ is even. By definition of even numbers, there exists an integer $$k$$ such that $$a = 2k$$.\
Going back to the original statement, we have $$2 = a^2 - 4ab = (2k)^2 - 4ab = 4k^2 - 4ab = 4(k^2 - ab)$$, or $$2 = 4(k^2 - ab)$$. Simplifying, we get $$1 = 2(k^2 - ab)$$, which means that $$1$$ is an even number, contradicting the fact that $$1$$ is odd.
We conclude that $$a^2 - 4ab \ne 2$$.
#### Scratchwork
When working with integers, it is often helpful to do cases where $$x < 0$$, $$x = 0$$, $$x > 0$$; or cases where $$x$$ is even, $$x$$ is odd. Let's try the latter.
#### Proof 2.
There are two cases: $$a$$ is even and $$a$$ is odd.
Case $$a$$ is even:\
By definition of even numbers, there exists an integer $$k$$ such that $$a = 2k$$. Then $$a^2 - 4ab = (2k)^2 - 4ab = 4k^2 - 4ab = 4(k^2 - ab)$$. So $$a^2 - 4ab$$ is a multiple of $$4$$. Since we know $$2$$ is not a multiple of $$4$$, $$a^2 - 4ab \ne 2$$.
Case $$a$$ is odd:\
By definition of odd numbers, there exists an integer $$k$$ such that $$a = 2k + 1$$. Then $$a^2 - 4ab = (2k + 1)^2 - 4ab = 4k^2 + 4k + 1 - 4ab = 2(2k^2 + 2k - 2ab) + 1.$$\
So $$a^2 - 4ab$$ is also an odd number. Since we know that $$2$$ is even, $$a^2 - 4ab \ne 2$$.
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