Suppose a and b are real numbers. Prove that if $a < b < 0$ then $a^2 > b^2$

Question

I'm reading the book "How to prove it" and this is one of the exercises. I'm trying to prove this statement and I was wondering if this counts as a proof.

Suppose a and b are real numbers. Prove that if $a < b < 0$ then $a^2 > b^2$

To start off we have:

$$(a < b < 0) \rightarrow (a^2 > b^2)\tag 1$$

We look at the contrapositive of this statement:

$$\lnot (a^2 > b^2) \rightarrow \lnot (a < b < 0)\tag 2$$

We look at the left side and write it as:

$$b^2 \leq a^2\tag 3$$

$$\frac{b^2}{a} \leq a\tag 4$$

The only way this can be true is if:

$$b^2 \leq a = b \leq \frac{a}{b}\tag 5$$

Which means

$$a \geq b\tag 6$$

Which means $b \not\gt a$

Therefore $$\lnot (a^2 > b^2) \rightarrow \lnot (a < b < 0)\tag 7$$ is true and we can conclude that

$$(a < b < 0) \rightarrow (a^2 > b^2)\tag 8$$

Does this count as a correct proof? If not can you point out my mistakes and how I possibly could have solved it? Thanks in advance.

Answer 1

Hm, I would just say that if $a < b < 0$, then we know that $|a| > |b|$, and thus $a^2 = (-|a|)^2 = |a|^2 > |b|^2 = (-|b|)^2 = b^2$ where the inequality stems from the monotonicity of the square function.

Answer 2

The step after you say "we look at the left side and write it as "..."" Is wrong. It should have been: $\lnot( a^2 \gt b^2)\equiv a^2\leq b^2$. Therefore, the proof is wrong.

Let me explain why that is wrong: $a \gt b$ means $(a\gt b)\land (a\neq b)$; therefore, when you negate it the conjunction becomes: $(a\lt b) \lor (a=b)$. Which is nothing but the implication of demorgan's law.

Instead of a direct proof I suggest you do an indirect proof:

Show: $P \implies Q$, by supposing $\lnot (P\implies Q)\equiv P \land \lnot Q$ and showing a contradiction:

Suppose: $(a\lt b\lt 0) \land a^2\leq b^2$

By quaring the first conjunct we get, $a^2\gt b^2\gt 0$, but we assumed in the second conjunct that $a^2\leq b^2$. Since both conjuncts are mutually contradictory. Our assumption that $\lnot (P\implies Q)$ is false, and therefore $P \implies Q$ is true.

Answer 3

A different and very basic proof is the following:

It holds that $a^2-b^2=(a-b)(a+b)>0$ since $a-b,a+b<0$.

Answer 4

We look at the contrapositive of this statement:

$$\lnot (a^2 > b^2) \rightarrow \lnot (a < b < 0)\tag 2$$

We look at the left side and write it as:

$$b^2 \leq a^2\tag 3$$

It's hard for me to work out what you're trying to do with this proof, but you seem to be trying to write $\lnot (a^2>b^2)$ here, if so this should say

$$b^2 \geq a^2\tag 3$$

You can either put the $a^2$ after the $b^2$ or you can put the inequality the other way around, but if you do both at the same time the two things cancel each other out. The only reason I can guess that you're trying to write the contrapositive is because you have $\leq$ rather than $<$ in your equation.

$$b^2 \leq a^2\tag 3$$

$$\frac{b^2}{a} \leq a\tag 4$$

Since we are given $a<0$, then when you divide by $a$ the inequality is reversed, so that should read

$$\frac{b^2}{a} \geq a\tag 4$$

The only way this can be true is if:

$$b^2 \leq a = b \leq \frac{a}{b}\tag 5$$

It is clear that $b^2>0>a$ so the first part is not true, then you have divided by a negative number again. Also the equals sign seems to be a substitute for $\implies$ (\implies) here.

Does this count as a correct proof? If not can you point out my mistakes and how I possibly could have solved it?

No it's not a correct proof at all. You might be able to rescue it if you can sort out all of the problems with division by negatives and putting the inequalities the wrong way around.

But you can prove this result by noting that $(a+b)(a-b)=\text{negative}\times\text{negative}=\text{positive}$ and $(a+b)(a-b)=a^2-b^2$.