If $ab>0$, then which one of the following must be true?

Question

Problem #$1$

If $ab>0$, then which one of the following must be true?

$(A)$ $a/b>0 $
$(B)$ $a-b>0 $
$(C)$ $a+b>0 $
$(D)$ $b-a>0 $
$(E)$ $a+b<0 $

Problem #$2$

If $x+z > y+z$ then which of the following must be true?

$(A)$ $x-z>y-z$
$(B)$ $xz>yz$
$(C)$ $x/z>y/z$
$(D)$ $x/2>y/z$
$(E)$ $2x+z>2y+z$

What are the main points to remember to solve these kinds of problems?

Answer 1

You are dealing with an ordered field $F$ if there is a set $P$ (the P corresponds with "positive") such that the $3$ sets: $$P \text{, }\{0\}\text{ and }-P$$ form a partition (i.e. the sets are not-empty, are disjoint and cover $F$).

Here $-P:=\{-x\mid x\in P\}$

This with:$$\forall x,y\in P[x+y\in P\text{ and }xy\in P]$$i.e. the positive set $P$ is closed under addition and multiplication.

An order $<$ on $F$ is induced by:$$x<y\text{ if }y-x\in P$$

Equipped with this knowledge problems like #1 and #2 can always be solved by you.

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Edit:

It is handsome to start with the observation that $1\in P$, which is no extra knowledge, but is something that can be proved on the knowledge mentioned above (so is no extra knowledge).

Proof:

Suppose that $1\notin P$.

Then - because also $1\notin\{0\}$ we must have $1\in-P$ or equivalently $-1\in P$.

But $P$ is closed under multiplication allowing us to conclude that $1=(-1)(-1)\in P$ and a contradiction is found, so we conclude that $1\in P$.

Answer 2

What are the main points to remember to solve these kinds of problems?

I'll answer this by sharing my thought process when solving these.

#1

If $ab > 0$, then the product of those two numbers is positive. That means either $a$ and $b$ are both positive or both negative. If I have two numbers with the same sign (i.e. they are both positive or both negative), can I say anything definitively about their sum, difference, or quotient?

Final hint: If I have two numbers with the same sign (i.e. they are both positive or both negative), what can I say about their quotient?

#2

If $x+z>y+z$, then $x>y$. Can this inequality be manipulated to match one of the answer choices?

Here's a narrative hint: suppose we are both in a race, and I am ahead of you. If we both walk backwards the same distance, what is my distance ran compared to yours?