$$\sqrt{x^2}=\sqrt{25} \;\;\;\;\;\;\;\Longleftrightarrow\;\;\;\;\;\;\; |{x}|=5 \;\;\;\;\;\;\;\Longleftrightarrow\;\;\;\;\;\;\; x=-5\;\;\;\;\;\;\text{or}\;\;\;\;\;\;x=5$$

This is the reason why we must never forget to include the absolute value when taking the square root of squared variables!

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Theorem.

Taking the square root of both sides of a squared less than b squared

If $a\in\mathbb{R}$ and $b\in\mathbb{R}$, then:

$$a^2\lt{b}^2 \;\;\;\;\;\implies\;\;\;\;\; |a|\lt{|b|}$$

Intuition. In order to prove this, we can show that the result holds for all 4 possible cases:

$a$ and $b$ are both positive.
$a$ is positive and $b$ is negative.
$a$ is negative and $b$ is positive.
$a$ and $b$ are both negative.

Instead of sticking with abstract variables $a$ and $b$, let's make things more concrete by using actual numbers. For the case when $a$ and $b$ are both positive, consider the following inequality:

$$3^2\lt5^2 \implies |3|\lt|5|$$

For the case when $a$ is positive and $b$ is negative:

$$(3)^2\lt(-5)^2\implies |3|\lt|-5| $$

Notice how the absolute value is required in this case.

For the case when $a$ is negative and $b$ is positive:

$$(-3)^2\lt(5)^2\implies |-3|\lt|5| $$

Here, the absolute value is not required since $-3\lt5$ is still true, but we can make a stronger claim by including the absolute $\vert{-3}\vert\lt5$. This point is important because many mathematical proofs use this stronger claim.

Finally, for the case when $a$ and $b$ are both negative:

$$(-3)^2\lt(-5)^2\implies |-3|\lt|-5| $$

We see that in all 4 cases, our theorem holds true!

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Theorem.

Taking the square root of both sides of a squared less than b

If $a\in\mathbb{R}$ is any number and $b\in\mathbb{R}$ is a non-negative number, then:

$$a^2\lt{b} \;\;\;\;\;\implies\;\;\;\;\; |a|\lt{\sqrt{b}}$$

Proof. Recall from the previous theorem that:

$$a^2\lt{c}^2 \;\;\;\;\;\implies\;\;\;\;\; |a|\lt{|c|}$$

Where a is any number and c is a non-negative number. Let's define a new variable $b=c^2$. Since $b$ is the square of $c$, we have that $b$ must also be a non-negative number. We substitute $b=c^2$ above to get:

$$a^2\lt{b} \;\;\;\;\;\implies\;\;\;\;\; |a|\lt{|\sqrt{b}|}$$

Again, because $b$ is a non-negative number, $\sqrt{b}$ must also be non-negative. This allows us to get rid of the absolute value:

$$a^2\lt{b} \;\;\;\;\;\implies\;\;\;\;\; |a|\lt{\sqrt{b}}$$

Remember, $b$ is a non-negative number so we've managed to prove the theorem!

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Published by Isshin Inada

Edited by 0 others

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