1 | The important equation. |
Prove that if $a$ is a real number, then $|a| = |-a|$.
Todo
Proof.
- Case 1: Suppose that $a \geq 0$. Blah blah blah. Therefore, $|a| = |-a|$.
- Case 2: Suppose that $a < 0$. Blah blah blah. Therefore, $|a| = |-a|$.
Sarah's contribution transcribed by Dr. Bagley
If $\exists a \in \mathbb{R}$, we can use Definition 5. It says the absolute value $|x|$ is either $x$ or $-x$. We can see that we get $|-a|$ when / if $x < 0$.
Using Definition 5,
- (c1) Suppose that $a\geq 0$, then $|a| = a$
- (c2) Suppose that $a < 0$, then $|a| = |-a|$
We can see that case 2 is true by Definition 5.
case 1:
Suppose a is a real number and is greater than or equal to 0.
By definition 5, |a| = a
Because a is greater or equal to 0, we know that -a is less than or equal to 0,
So, |-a| = - -a by definition 5. $
Is this correct? Because I switched the sign to see that -a is less than or equal to 0, then I had to look back at 'case 2' of definition 5.
Garrick's Contribution
Amin's Contribution
Case 1:
Let $$ a > 0 $$
then $-a$ < 0, by definition 5, we have $$ |-a| = -(-a) = a = |a|. $$
Case 2:
Let $$ a < 0 $$
then $-a$ > 0, by definition 5, we have $$ |-a| = -a = |a|. $$
Therefore for all all a E R we have $|-a| = |a|$ if we combine the two cases.