Chapter 1 · Algebraic Proofs · Concept 4
1-11 Proving Equalities (2)
The equalities in lesson 1-10 were true for every value of every letter. These ones are not. They are only true when something else is true first — and using that something is the whole method.
Guiding question
How do you prove an equality that is only true when a condition holds?
Learning outcome. Prove the validity of algebraic equalities and identities in conditional cases.
The method
Take the condition, write one letter in terms of the others, and put it in.
If x + y = 1, then write y = 1 − x and substitute it, so that y disappears.
Once one letter is gone, everything is in terms of the others and it becomes an ordinary lesson-1-10 proof: tidy each side on its own until they match.
This is the book's Warm Up. It is not true for any three numbers — only when they add to zero. Try it both ways.
a² + b² + c² = −2(ab + bc + ca)
left = 14 · right = -22
the equality is false here
a + b + c = 6 — the condition is not satisfied
Move the sliders with the box unticked and it breaks almost everywhere. Tick the box and it holds no matter what you do. That is what a condition is for — the equality is not true in general, only on the values the condition allows.
If a + b + c = 0, prove that a2 + b2 + c2 = −2(ab + bc + ca)
- 1
Use the condition to get rid of one letter.
from a + b + c = 0, c = −a − b …… (i)Any of the three would do. Choosing c is the tidiest because it appears least.
Work with a partner. Do these together, then talk about the answers. Decide which letter to get rid of before you write anything.
Prove the following equalities.
Three numbers a, b, c, none of them zero, satisfy a + b + c = 0. Prove that whichever such numbers you pick, this sum always comes to exactly −3:
a/(b+c) + b/(c+a) + c/(a+b) = −3
Use the condition — do not just mention it.
Writing "since a + b + c = 0" and then proving nothing with it gets no marks. It has to become c = −a − b and go into the working.
Still one side at a time.
The rule from lesson 1-10 has not gone away. Substitute into the left, tidy it, then start again with the right.
Watch (−a − b)².
It equals (a + b)² = a² + 2ab + b², all positive. Squaring removes the minus, and forgetting that is the commonest slip on page 26.
Choose the letter that appears least.
You may replace any of them. Picking the one that shows up fewest times means fewer substitutions and fewer chances to slip.
Lesson 1-12 does the same for conditions written as ratios.
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