Inequality proof, why isn't squaring by both sides permissible?

Question

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

I understand that the normal way to prove this is to multiply $a < b$ by $a$ and then by $b$ and then combine to get $a^2 < ab < b^2$. However, what I don't understand is why can't you just square both sides? Is that breaking any rule?

Likewise when proving: If $0 \le a < b$, and $a^2 < b^2$, then $a < b$. Why isn't taking the square root of both sides done?

I apologize if my tags are incorrect, but i wasn't sure what to put it under.

Answer 1

Generally $$a>b \implies f(a)> f(b), \forall \text{ strictly increasing f(x)}$$

Your statement is just for $x^2$ which is true for $\Bbb R^+$. You have to state this fact while writing it down.

Your proof is also correct...like M.B. replied

Answer 2

Short answer: We can't simply square both sides because that's exactly what we're trying to prove: $$0 < a < b \implies a^2 < b^2$$

More somewhat related details: I think it may be a common misconception that simply squaring both sides of an inequality is ok because we can do it indiscriminately with equalities. Let's take an example: $$x = 2$$ We know we can square both sides to get: $$x^2 = 4$$ But why can we square both sides? It's because squaring both sides is really just a special case of "multiplying both sides of the equality by the same thing." I think most of us are comfortable with the fact that if we have an equation such as $$\dfrac{2y}3 = \dfrac83$$ then to solve it we multiply both sides by $3/2$: \begin{align*} \frac32 \cdot \dfrac{2y}3 &= \dfrac83 \cdot \frac32\\[0.3cm] y &= 4 \end{align*} Well, that's exactly the same concept we used when we squared both sides of $x=2$. Let's take $x=2$ and multiply both sides by $2$: \begin{align*} x &= 2\\ 2 \cdot x &= 2 \cdot 2 \end{align*} On the left-hand side of the last line above, we can replace the $2$ with a $x$ because we already know that $x$ and $2$ are the same thing. So then: \begin{align*} x &= 2\\ 2 \cdot x &= 2 \cdot 2\\ x \cdot x &= 2 \cdot 2\\ x^2 &= 4 \end{align*} I'll reiterate the main point for emphasis: Squaring both sides here worked because we already know that $x$ and $2$ are equal. This is exactly why we can't do the same thing with inequalities in general. If we have $x < 2$ then we can't square both sides, because squaring both sides is a special case of multiplying both sides by the same thing, and if $x < 2$ then $x$ and $2$ are not equal.

If $x < 2$ then $x^2 < 4$ is true if $x$ is, for example, $1$, and $x^2 < 4$ is false if $x$ is, for example, $-3$. So if we want to square both sides of $x < 2$ and still have a true inequality, then we need the additional restriction that $x > 0$. (Actually $x \ge 0$ is sufficient.)

But why is this restriction enough to make squaring both sides of the inequality ok? That's exactly what the original question wants you to answer. Yes, by the time you get to a math course involving proofs you should be very familiar with the fact that $0 < a < b \implies a^2 < b^2$, but now you actually have to explain why it's true. And the proof you outlined in your question (first multiply both sides by $a$, then by $b$) is exactly how we do that.

Answer 3

"Likewise when proving: If $0\leq a<b$, and $a^2<b^2$, then $a<b$. Why isn't taking the square root of both sides done?"

There is an intermediate step to this.

$$a^2<b^2$$

$$\sqrt{a^2}<\sqrt{b^2}$$ ... notice that the square is inside the square root. Recall that this is equal to the absolute value. Square root of a number squared is equal to the absolute value of that number

$$ |a| < |b| $$

... This is where it gets hairy, and depends case-to-case.

a) Here, both values are non-negative, ... the abs of which are themselves: $|a| = a, |b| = b$

$$ a < b $$

b) If both are non-positive, $|a| = -a, |b| = -b$, The abs resolves to: $$ -a < -b $$

$$ b < a $$ ... I prefer not to "multiply by -1" , but rather transpose a to the right and b to the left.

$$ a > b $$

c,d) If they have different signs, The abs conditionally resolves to

$$ a < -b \quad if \quad a\geq 0,b\leq 0 $$

$$ -a < b \quad if \quad a\leq 0,b\geq 0 $$

WHICH DO NOT GIVE ANY RELATIONSHIPS BETWEEN a AND b,

...but rather with the inverse of the other.

I always live by my math professor's advice: If you can't specify what rule you're using on your equations, don't use it.

CONCLUSION: I hope this answers your question, as to "why you cant",

... and better yet, "why you can in some cases."

Answer 4

This answer is very late, but this is another way to view the problem:

We are trying to prove $a^2 > b^2$, or $a^2-b^2>0$.

Using the identity: $a^2-b^2$ = $(a+b)(a-b)$,

we are equivalently trying to prove:

$(a+b)(a-b) > 0 $

Since $0<a<b$,

$(a+b) > 0 $.

Now, the LHS is only positive if $a-b>0$ or $a>b$ which has been our assumption from the beginning, where $a,b>0$

And the proof falls into place. The only problem here is that we are starting from what we want to prove and building from that, but one could rewrite the proof so that doesn't happen.