Inequality and absolute value proof

Similarly, in the second proof, you don't "take the absolute value of both sides". You replace a and b by |a| and |b| in the inequality and that gives |a|- |b|\le |a- b|.In general, if you want to get |a- b| on the left, you have to subtract something from |a| and |b|. The two methods you give are the two ways to do that.
  • #1
EV33
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Homework Statement


prove that llal-lbll[tex]\leq[/tex]la-bl

Homework Equations



Triangle inequality
lx+yl[tex]\leq[/tex]lxl+lyl

The Attempt at a Solution



Let a=(a-b)+b
By using the triangle inequality we get
lal-lbl[tex]\leq[/tex]la-bl

Then from here I am not sure what I can do. I would like to say on the left hand side that I can use the triangle inequality again by taking the absolute value of the left side and saying that I have the absolute value of a plus -absolute value of b.

llal+(-lbl)l[tex]\leq[/tex]lal-lbl[tex]\leq[/tex]la-bl

My second thought it instead of doing that I could possible just take the absolute value of both sides and then I would get what I was trying to prove because taking the absolute value of the right side wouldn't change anything, but taking the absolute value of the left side would change it exactly the way I need it to be.

I was wondering if either one of those Ideas is legitament because I am really not sure about them. If neither one is could someone point me in the right direction in proving this. Thank you for your time.
 
Last edited:
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  • #2


Sorry I will have to edit this. I don't know what happened.
 
  • #3
fixed it
 
  • #4
In general taking absolute value of both sides does not preserve an inequality.

For example:
-2 < -1
|-2| > |-1|

So that doesn't work.

Your first method seems to work much better.
 
  • #5
So the first method works for sure? It seems reasonable to me but someone I talked to seemed like they thought it might not work. So I just want to make sure.

Thank you.
 
  • #6
Yes, your first proof works. However, you shouldn't say "let a= (a- b)+ b". You don't need to "let" that be true- it is! What you mean to say is "in [itex]|x+ y|\le |x|+ |y|[/itex], let x= a- b and y= b". Then x+ y= (a- b)+ b= a so [itex]|a|\le |a- b|+ |b|[/itex] and, subtracting |b| from both sides, [itex]|a|- |b|\le |a- b|[/itex].
 

Related to Inequality and absolute value proof

1. What is the definition of inequality?

Inequality is a mathematical concept that describes a relationship between two values or expressions, where one is greater than or less than the other.

2. What is an absolute value?

An absolute value is the distance of a number from zero on a number line. It is always a positive value.

3. How do you prove an inequality using absolute values?

To prove an inequality using absolute values, you must show that the absolute value of one side is greater than or less than the absolute value of the other side.

4. What is the difference between an absolute value and a regular value?

The absolute value of a number is always positive, while a regular value can be either positive or negative. In other words, the absolute value tells us the distance of a number from zero, regardless of its sign.

5. Can you give an example of proving an inequality using absolute values?

Yes, for example, to prove that |a+b| > |a| + |b|, we can consider two cases: when both a and b are positive, and when one of them is negative. In both cases, the absolute value of the left side is greater than the absolute value of the right side, thus proving the inequality.

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