Math Is Fun Forum

The rules for combining inequalities are subtler than the rules for equalities. While
a = b and a′ = b′ imply all three of a + a′ = b + b′a − a′ = b − b′ aa′ = bb′, on the other hand
(1) a < b and a ′ < b′ imply only the first of
(2) a + a′ < b + b′
(3) a − a′ < b − b′
(4) aa′ < bb′
because the other two are not true in general.

However, the other two can be salvaged, meaning an important special case or cases can be proved true, or some variation of what is stated can be proved true. Students tend to regard statements about inequalities as axiomatic so basic that they can’t be proved and must be assumed. Not so! Given the definition of “<” below, all the properties of inequality can be reduced to properties of positive and negative numbers.

Here is an example:
Definition. We write a < b if (and only if) b = a+ p, where p is a positive number.
Theorem. If a < b and b < c, then a < c.
Proof: By definition of <, we are given that b = a + p1 , where p1 is positive. We are also given by definition that c = b + p2, where p2 is positive. Therefore c = b + p2 = (a + p1) + p2 = a + (p1 + p2).
Now the sum of two positive numbers is positive (this is the one fact about positive and negative numbers we use in this example). Thus by definition of < applied to a and c, we conclude that a < c.
a) Assuming display (1) prove inequality (2).
b) Assuming display (1) disprove inequality (3) and then salvage.
c) Assuming display (1) disprove inequality (4) and then salvage.
In your proofs, do not assume any facts about inequalities unless you have already proved them, and all proofs of inequality facts should rely on facts about positive and negative numbers, using the definition of < as in the example theorem above.

I can easily prove/disprove a, b, and c. But I need help salvaging b and c.