Proving |a|=|-a|: Using Cases and Triangle Inequality"

[stunner5000pt]

Homework Statement

Prove that |a|=|-a|

Relevant Equations

$|a|= a,$ if $a \geq 0$ and
-a, if $a \leq 0$

Problem Statement: Prove that |a|=|-a|
Relevant Equations: $|a|= a,$ if $a \geq 0$ and
-a, if $a \leq 0$

Somewhat stumped on where to start...

i know that we need to use cases. If we consider $a\geq 0$, then are we allowed to use the fact that $|-a|=|-1|\cdot|a| = |a|$?

This is from Spivak so in the P1 to P12, and triangle inequality, the above has not been developed.

What about using the triangle inequality? But how would we separate $|a|-|-a|$?

Any input would be greatly appreciate! Thank you in advance

 

[YoungPhysicist]

Absolute value represents the distance between the number’s( regardless of positive or negative) distance to 0 on the number line.

Since $a$ and $-a$ are same distance, which is $a$ away from 0, their absolute value is the same.

 

[ali PMPAINT]

Since $a$ and $-a$ are same distance,.

Yes, but you should prove it first, you can't use a statement to prove itself.

 

[PeroK]

Problem Statement: Prove that |a|=|-a|
Relevant Equations: $|a|= a,$ if $a \geq 0$ and
-a, if $a \leq 0$

Problem Statement: Prove that |a|=|-a|
Relevant Equations: $|a|= a,$ if $a \geq 0$ and
-a, if $a \leq 0$

Somewhat stumped on where to start...

i know that we need to use cases. If we consider $a\geq 0$, then are we allowed to use the fact that $|-a|=|-1|\cdot|a| = |a|$?

If Spivak has already proved that $|ab| = |a||b|$, then of course you can use that result here. But, if not then of course you cannot use this result.

When you are learning rigorous maths like this it's important not only to start to think differently but also to remember what you have already proved (and what you know to be true but haven't proved formally yet).

Another key point in always to go back to the definition of things. If you want to prove something formally about $||$ before you've proved anything else about it, then you must start with the definition.

 

[SammyS]

Problem Statement: Prove that |a|=|-a|
Relevant Equations: $|a|= a,$ if $a \geq 0$ and
-a, if $a \leq 0$

Problem Statement: Prove that |a|=|-a|
Relevant Equations: $|a|= a,$ if $a \geq 0$ and
-a, if $a \leq 0$

Somewhat stumped on where to start...

i know that we need to use cases. If we consider $a\geq 0$, then ...

... are we allowed to use the fact that $|-a|=|-1|\cdot|a| = |a|$?​

I doubt that you can use that. However, I suspect that you can show:

If $a\geq 0$, then $-a \le 0 \,.$​

Am I right?

If so, then what is $\left|-a\right|$ for this case?

 

[stunner5000pt]

ok let's try this again, this time, let's use cases like hinted above

Consider $a \geq 0$ , then $|a| = a$
and $|-a| = a$

We can say the second statement is true because absolute value of a negative number is the negative of the number (the definition of absolute value)

Then the version where $a < 0$ follows.

how would this work as a proof?

 

[SammyS]

ok let's try this again, this time, let's use cases like hinted above

Consider $a \geq 0$ , then $|a| = a$

Prior to asserting the following, it seems to me that you need to show that $(-a)$ is truly less than or equal to zero.

and $|-a| = a$

Then using the given definition of absolute value, you literally get that $|-a| = -(-a)$ .

From this it's no problem to show that $|-a| = a$ .

Thus for this case, in which $a\ge0$, you have that both $|a| = a, \text{ and } |-a| =a$ .

We can say the second statement is true because absolute value of a negative number is the negative of the number (the definition of absolute value)

Then the version where $a < 0$ follows.

how would this work as a proof?