Inequality properties by Dr. Wiggins
MAM1031F, UCT
What is a property? In mathematics, a property is any characteristic that applies to a given set.
For example real numbers, the rules governing the relationship and interaction of real numbers with
each other. The basic ones are commutative, associative, distributive. Google, if you don’t know
what these words mean.
What is a direct proof ? In mathematics, a direct proof is a way of showing the truth of a given
statement by a straightforward combination of established facts, usually axioms, existing lemmas and
theorems, without making any further assumptions using solid mathematical logic. If the statement
is “If P then Q”, then the method of the proof is to takes an original statement P , which we assume
to be true, and use it to show directly that another statement Q is true.
Here is 6 basic inequality properties for real numbers:
Suppose a, b, c, d are real numbers.
(P1) If a ≤ b and b ≤ c then a ≤ c
(P2) If a ≤ b then a + c ≤ b + c
(P3) If a ≤ b and c ≤ d then a + c ≤ b + d
(P4) If a ≤ b and c ≥ 0 then ac ≤ bc
(P5) If a ≤ b and c ≤ 0 then ac ≥ bc
1
1
(P6) If 0 < a ≤ b then ≥
a
b
1. Prove P3 is a consequence of P1 and P2. That is if P1 and P2 holds then P3 is true.
2. Prove if x < 0 then −x > 0. Make sure you use P2 in your argument.
3. Use problem 2, P2 and P4 to prove P5.
Please TRY THEM without looking at the solutions. Only after struggling for at least a hour, go to
the next page for the solution.
Success in math does not depend on how many answers you know, but by what you do
when you don’t know the answer.
Are you sure you want to see the answers?
1. Prove P3 is a consequence of P1 and P2. That is if P1 and P2 holds then P3 is true.
Solution: Strategy P3 is the statement: If “a ≤ b and c ≤ d” then “a + c ≤ b + d”. So we
assume the first statement and want to deduce the second statement.
Assume that a ≤ b and c ≤ d.
Recall P2 says if p ≤ q then p + r ≤ q + r. So if a ≤ b by adding c you get
a+c≤b+c
(1)
Recall P2 says if p ≤ q then p + r ≤ q + r. Using c ≤ d by adding b you get
b+c≤b+d
(2)
Recall P1 says if p ≤ q and q ≤ r then p ≤ r. Now we have from equation (1) that
a + c ≤ b + c. We got from equation (2) that b + c ≤ b + d.
Hence by P1 we get a + c ≤ b + d as desired completing this direct proof.
2. Prove if x < 0 then −x > 0. Make sure you use P2 in your argument.
Solution: Strategy We want to show: If “x < 0” then “−x > 0. So we assume the first
statement and want to deduce the second statement. And we have to use property P2 and
tell the reader where in the argument we are using P2.
Assume x < 0.
Recall P2 says if p < q then p + r < q + r. So if x < 0 by adding −x you get
x + (−x) < 0 + (−x)
0
−x
< −x
>
0
Here we used that m < n is equivalent to n > m. So assuming by using the statement
“x < 0” and a mathematical argument using P2, we deduced that “−x > 0”.
3. Use problem 2, P2 and P4 to prove P5.
Solution: Strategy We want to show: If “a ≤ b and c ≤ 0” then “ac ≥ bc. So we assume
the first statement and want to deduce the second statement. And we have to use property
P2,P4 and the result from problem 2. So ensure you tell the reader where in the argument
you use these established results.
Assume a ≤ b and c ≤ 0.
Using problem 2, if c ≤ 0 then −c ≥ 0. Recall P4 says that if p ≤ q and r ≥ 0 then pr ≤ qr.
So if a ≤ b and −c ≥ 0 then −ac ≤ −bc.
Again, recall that P 2 states if p ≤ q then p + r ≤ q + r. So −ac ≤ −bc we add ac + bc to
both sides and get
−ac + (ac + bc) ≤
−bc + (ac + b)
bc ≤ ac
ac ≥ bc
The last line, used the fact that m ≤ n is equivalent to n ≥ m. So we deduced the statement
ac ≥ bc as desired.
Remark: A trap is to use −1 < 0. You cannot use −1 < 0 since it wasn’t part of the given
inequality properties.
If you spot maths errors, spelling errors, etcetera. Please report them to harry.wiggins@uct.ac.za.
Or email me if you have a different elegant solution.