Prove $a^2 = b^2$ implies that $|a| = |b|$

Question

STRUGGLING TO UNDERSTAND TRICHOTOMY...

This question is from Peter Eccles' excellent book - An Introduction to Mathematical Reasoning.

All variables and functions are concerning properly defined real numbers.

It is given that we have already proven the converse of the featured question.

The answer in the book is as follows:

Suppose that $a^2 = b^2$ and, for contradiction, that $|a|\neq|b|$. Then by trichotomy, either $|a|<|b|$ in which case $a^2 < b^2$ giving a contradiction, or $|a| > |b|$, in which case $a^2 > b^2$ giving a contradiction. Since we necessarily obtain a contradiction our assumption that $|a|\neq|b|$ must be false and so $|a| = |b|$ as required, thus proving that $a^2 = b^2$ implies $|a|=|b|$.

I am familiar with proof by contradiction, but something is rankling. Does the LHS allow us to speculate about the RHS trichotomously?

Is it true that, given that $g(a) = g(b)$ implies $f(a) = f(b)$ then $f(a) = f(b)$ implies $g(a) = g(b)$ providing $g(a)$ is neither greater than nor less than $g(b)$?

Could we in fact dispense with having proved the converse case altogether so that we are not given $g(a) = g(b)$ and still conclude the same? (in terms of the original question concluding that $|a| = |b|$)

I think the difficulty I am having with this is to do with the mathematical meaning of implication, specifically the aspect of the truth table whereby $p$ is false, $q$ is true but $p$ still implies $q$.

It seems like the 'correspondence is not causation' phrase one hears in connection with statistics could also be applicable here.

In a closed system such as the reals, all elements simply must be somewhere on the number line and simply must have a relative position to every other point.

Answer 1

The proof is a "usual" proof by contradiction: sketch of the proof.

The proof starts from: $a^2=b^2$.

$a^2$ is defined as $a \cdot a$ and thus it is a real number (real numbers are a field and a field is an algebraic structure closed under two operations called addition: $+$ and multiplication: $\times$ or $\cdot$).

Absolute value is a real function defined on reals, i.e. it maps real to real.

Trichotomy applies to real numbers because they for an ordered field, where $\le$ is the ordering relation.

Trichotomy follows from the following definition:

$a < b \text { iff } \lnot (b \le a)$.

The defintion implies that: $a < b \text { iff } (a \le b \text { and } a \ne b)$.

[Easy check: from $a < b$, applying the definition, we have that: $\lnot (b \le a)$ that implies (using the total order axioms): $(a \le b)$. But if $a=b$, then also $b \le a$, contradicting the definition; thus, we have also: $\lnot (a=b)$].

These are the basic ingredients.

$|a|$ and $|b|$ are reals, and thus we can apply Trichotomy to them. The disjunction:

$|a|=|b| \lor |a|<|b| \lor |a|>|b|$,

can be rewritten as:

$|a| \ne |b| \to (|a|<|b| \lor |a|>|b|)$.

Thus, assuming $|a| \ne |b|$, we apply Disjunction elimination to $(|a|<|b| \lor |a|>|b|)$.

Having found a contradiction under both disjuncts, we conclude that the disjunctions implies a contradiction.

Now, in order to show that e.g. $|a|<|b|$ implies a contradiction, under the assumption that $a^2=b^2$, we need some steps using further properties of reals.

From $|a|<|b|$, using the order axiom: if $x ≥ 0$ and $y ≥ 0$ then $xy ≥ 0$, we have: $0 \le |a|,|b|$.

In addition, we need the related property that: if $x ≥ y ≥ 0$ and $a \ge 0$, we have: $ax ≥ ay ≥ 0$.

This implies:

$0 \le |a| \cdot |a| < |a| \cdot |b| < |b| \cdot |b|$.

But we have that $a^2= a \cdot a$. Using the property of absolute value: $|a \cdot b|=|a| \cdot |b|$, we have that $|a^2|=|a| \cdot |a|$, and the same for $b$.

And it's (quite) done.

From case $|a| < |b|$ we have found that $|a| \cdot |a| < |b| \cdot |b|$.

From assumption $a^2=b^2$ we have found that: $|a| \cdot |a| = |b| \cdot |b|$.

Using properties of equality we get:

$|a| \cdot |a| < |a| \cdot |a|$,

contradicting the fact that, for total orders, the strict total order implies that:

if $a < b$, then $a \ne b$.