−4 and 4 are different numbers, yet both sit exactly four steps from the origin. Absolute value measures only that distance and forgets the direction. Everything in this note grows out of that one idea, including the triangle inequality, one of the proofs the boards ask for most.
1. Absolute value is a distance
The distance on the number line between the origin (0) and the point of a number is that number's absolute value. The absolute value of −4 is 4, and of 3 it is 3. We write |−4| = 4 and |3| = 3.
A distance is never negative, so the rule comes in three parts: if x > 0 then |x| = x; if x = 0 then |x| = 0; if x < 0 then |x| = −x. In short, |x| = x for x ≥ 0 and −x for x < 0. Every real number except zero has a positive absolute value, and zero's is zero. Move x below and watch which part of the rule is in force.
measure the distancedrag the handle
How far is x from 0?
The square root says the same thing: for every x ∈ ℝ, |x| = √(x²). That is because |x|² = x², and since |x| ≥ 0 the negative root is thrown out. So |5| = 5, |−5| = −(−5) = 5 and |0| = 0.
Two small facts follow: for every a, −|a| ≤ a ≤ |a|, and |a| ≥ a. Prove them in two cases. If a ≥ 0, then −|a| ≤ 0 ≤ a = |a|. If a < 0, then −|a| = a < 0 < −a = |a|, so here |a| = −a > a.
2. Products and the triangle inequality
For all a, b ∈ ℝ, |a|² = a² = |−a|², and |ab| = |a||b|. Proof: |ab|² = (ab)² = a²b² = (|a||b|)². Both sides are non-negative, so taking square roots gives |ab| = |a||b|.
Sums do not give an equality, they give an inequality. The triangle inequality: for all a, b ∈ ℝ, |a + b| ≤ |a| + |b|. Think of a and b below as arrows. Walk both the same way and the whole walk is distance gained; walk them opposite ways and you come back some of the way.
two arrowschange a and b
When does |a + b| reach |a| + |b|?
The proof goes through squares. (|a| + |b|)² = a² + 2|ab| + b². Since |ab| ≥ ab, this is ≥ a² + 2ab + b² = (a + b)². So |a + b|² ≤ (|a| + |b|)². Both sides are non-negative, so take square roots: |a + b| ≤ |a| + |b|. Put the steps in order yourself below.
build the proofsquare first, root last
Proof of the triangle inequality
Start by squaring the right-hand side and expanding it.
Yes. Which line follows from it?
Not yet. This line needs the one before it: expand, then |a|² = a² and |a||b| = |ab|, then |ab| ≥ ab.
Proof complete. The square root in the last step is allowed because both sides are non-negative.
Two more follow from it. Put −b in place of b in the triangle inequality: |a + (−b)| ≤ |a| + |−b| = |a| + |b|, that is |a − b| ≤ |a| + |b|.
And a bound from the other side: |a − b| ≥ ||a| − |b||. From |a| = |(a − b) + b| ≤ |a − b| + |b| we get |a − b| ≥ |a| − |b|. In the same way |b| ≤ |a − b| + |a| gives |a − b| ≥ −(|a| − |b|). Together, |a − b| ≥ ||a| − |b||.
The formula list also has |ab| ≥ ab and |a/b| = |a|/|b| (b ≠ 0). For positive a and b, |a + b| = a + b = |a| + |b|, so equality holds. But |a − b| = |a| + |b| does not: with a = 3, b = 2 you get 1 ≠ 5. Now sort the true ones from the traps.
true or false?chip, then bin
For all a, b ∈ ℝ
Pick up a chip, then drop it in a box. One counter-example is enough to call it false.
Right. Next one.
Not yet. Try a = 2, b = −3 or a = 3, b = 2.
All in place. The traps: for a = 3, |a|² = 9 = a², not bigger. For a = 2, b = −3, |a − b| = 5 while ||a| − |b|| = 1. And for positive a, b, |a − b| is in fact smaller than |a| + |b|.
3. From the inside out
When one absolute value sits inside another, open the innermost bars first and the outer ones last. ||2 − 6| − 10 + |7 − 3|| = ||−4| − 10 + |4|| = |4 − 10 + 4| = |−2| = 2.
For a = 3, b = −7, c = −9, ||a − b| − c| = ||3 + 7| + 9| = |10 + 9| = 19. In the same way ||4 − 10| − |7 − 10| + |−3 − 2|| = |6 − 3 + 5| = 8, and |−11 + 7| − |2 − |−3|| + |−1 − 3| = 4 − 1 + 4 = 7. Pick one below and open it a layer at a time.
peel the barsstart inside
Evaluate the absolute values
4. Check yourself
Three questions in the exam's style. The first is about the properties.
check yourselfpick one answer
Question 1
For a, b ∈ ℝ, which of the following is correct?
|a + b| ≥ |a| + |b| does not fit here; the right answer is |a − b| ≤ |a| + |b|. |a − b| ≤ |a| + |b| is the book's property; the others are reversed or false.
|a − b| ≥ |a| + |b| does not fit here; the right answer is |a − b| ≤ |a| + |b|. |a − b| ≤ |a| + |b| is the book's property; the others are reversed or false.
Right. |a − b| ≤ |a| + |b| is the book's property; the others are reversed or false.
|a − b| ≤ ||a| − |b|| does not fit here; the right answer is |a − b| ≤ |a| + |b|. |a − b| ≤ |a| + |b| is the book's property; the others are reversed or false.
In the second, open the inner bars first.
check yourselfpick one answer
Question 2
What is the value of ||6 − 11| − 5 + |7 − 4||?
2 does not fit here; the right answer is 3. ||−5| − 5 + |3|| = |5 − 5 + 3| = 3.
Right. ||−5| − 5 + |3|| = |5 − 5 + 3| = 3.
5 does not fit here; the right answer is 3. ||−5| − 5 + |3|| = |5 − 5 + 3| = 3.
6 does not fit here; the right answer is 3. ||−5| − 5 + |3|| = |5 − 5 + 3| = 3.
The last one goes the same way.
check yourselfpick one answer
Question 3
What is the value of ||−7 + 5| − |−6 + 3| + |11 − 7||?
2 does not fit here; the right answer is 3. |2 − 3 + 4| = 3.
Right. |2 − 3 + 4| = 3.
4 does not fit here; the right answer is 3. |2 − 3 + 4| = 3.
5 does not fit here; the right answer is 3. |2 − 3 + 4| = 3.
The main points of the note, on cards.
cut from the notetap to turn over
Eight cards
QuestionWhat is the absolute value of a number on the number line?
AnswerThe distance between the origin (0) and the point of the number.
QuestionWhat is |x| when x < 0?
Answer−x (for example |−5| = 5).
QuestionHow is |x| written using a square root?
Answer|x| = √(x²).
QuestionHow is a related to −|a| and |a|?
Answer−|a| ≤ a ≤ |a|.
QuestionHow is |a + b| related to |a| + |b| (triangle inequality)?
Answer|a + b| ≤ |a| + |b|.
QuestionHow is |a − b| related to ||a| − |b||?
Answer|a − b| ≥ ||a| − |b||.
QuestionFor positive a, b why is |a + b| = |a| + |b| but |a − b| = |a| + |b| false?
AnswerFor positive a, b, |a + b| = a + b = |a| + |b|; but a − b subtracts, so |a − b| < |a| + |b| (for a = 3, b = 2: 1 < 5).