My Math Forum Inequality proof!!

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 March 4th, 2013, 09:01 PM #1 Member   Joined: Sep 2012 Posts: 69 Thanks: 0 Inequality proof!! Today I was told that my proof was wrong!! The question was this! Let $a,b,x,y \in R$ and $r \in R^+$. Prove that if$|x-a| < \frac{r}{2}$ and $|y-b| < \frac{r}{2}$, then $|(x+y) - (a+b)| < r$. My proof was this!! $|x-a| + |y - b|=< \frac{r}{2} + \frac{r}{2}= r$. Since $|x-a| + |y - b| < r$ and r is positive real number, $|x-a| + |y - b| < 0$. $|x-a| + |y - b|= -(x-a) - (y-b) = -[(x-a) + (y-b)] = -[(x+y) - (a+b)] = |(x+y) - (a+b)| < r=$. Can anyone comment on this?? Thanks
March 5th, 2013, 02:27 AM   #2
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Re: Inequality proof!!

Quote:
 $|x-a| + |y - b| < 0$
What is it? It is just not true.

Quote:
 latex]|x-a| + |y - b| = -(x-a) - (y-b)[/latex]
Why did you replace the absolute value with negative value? It is not a correct step.

 March 5th, 2013, 04:38 AM #3 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: Inequality proof!! Let $X, Y \in \mathbb{R}$, what do you know about relation $T$ in $|X|+|Y|\, T \,|X + Y|$? (You might generally have seen using R for relations, but it would be slightly ambigious now I think.)
March 5th, 2013, 01:12 PM   #4
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Re: Inequality proof!!

Quote:
 Originally Posted by eChung00 Today I was told that my proof was wrong!! The question was this! Let $a,b,x,y \in R$ and $r \in R^+$. Prove that if$|x-a| < \frac{r}{2}$ and $|y-b| < \frac{r}{2}$, then $|(x+y) - (a+b)| < r$. My proof was this!! $|x-a| + |y - b|=< \frac{r}{2} + \frac{r}{2}= r$. Since $|x-a| + |y - b| < r$ and r is positive real number, $|x-a| + |y - b| < 0$. $|x-a| + |y - b|= -(x-a) - (y-b) = -[(x-a) + (y-b)] = -[(x+y) - (a+b)] = |(x+y) - (a+b)| < r=$. Can anyone comment on this?? Thanks
Basic proof:
(x+y) - (a+b) = (x-a) + (y-b)
|(x-a) + (y-b)| ? |x-a| + |y-b| < r

March 5th, 2013, 05:19 PM   #5
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Re: Inequality proof!!

Quote:
Originally Posted by csak
Quote:
 $|x-a| + |y - b| < 0$
What is it? It is just not true.

[quote:3nbm04il]latex]|x-a| + |y - b| = -(x-a) - (y-b)[/latex]
Why did you replace the absolute value with negative value? It is not a correct step.[/quote:3nbm04il]

Quote:
 [quote:3nbm04il]latex]|x-a| + |y - b| = -(x-a) - (y-b)[/latex]
Why did you replace the absolute value with negative value? It is not a correct step.[/quote:3nbm04il][/quote]
I used the definition of the absolute value. $|x|= x\ if\ x \geq 0 \ or\ -x\ if\ x<0=$

March 5th, 2013, 08:31 PM   #6
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Re: Inequality proof!!

Quote:
 Originally Posted by eChung00 I used the definition of the absolute value. $|x|= x\ if\ x \geq 0 \ or\ -x\ if\ x<0=$
This is not true in the way you used it. By the definition of absolute value, $\text{abs} \; : \; \mathbb{C} \to \mathbb{R}^+$. You'll see that you are simply defining the function $f(x)= x$ if you look at your application in the OP

March 5th, 2013, 09:29 PM   #7
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Re: Inequality proof!!

Quote:
Originally Posted by mathbalarka
Quote:
 Originally Posted by eChung00 I used the definition of the absolute value. $|x|= x\ if\ x \geq 0 \ or\ -x\ if\ x<0=$
This is not true. By the definition of absolute value, $\text{abs} \; : \; \mathbb{C} \to \mathbb{Z}^+$. You are simply defining the function $f(x)= x$
I don't get it!! Can you explain it in more detail??
I copied the definition of absolute value exactly from my text book.

March 5th, 2013, 09:54 PM   #8
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Re: Inequality proof!!

Quote:
 Originally Posted by eChung00 I don't get it!! Can you explain it in more detail?? I copied the definition of absolute value exactly from my text book.
You copied it exactly but didn't understood it. The definition is correct, $|x|= -x$ for x<0 but that doesn't mean that $|-3|= -3$ but it's $|-3|= 3$. You used this in the other way without changing the sign by multiplying -1 on the RHS.

March 6th, 2013, 06:39 AM   #9
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Re: Inequality proof!!

Quote:
Originally Posted by mathbalarka
Quote:
 Originally Posted by eChung00 I don't get it!! Can you explain it in more detail?? I copied the definition of absolute value exactly from my text book.
You copied it exactly but didn't understood it. The definition is correct, $|x|= -x$ for x<0 but that doesn't mean that $|-3|= -3$ but it's $|-3|= 3$. You used this in the other way without changing the sign by multiplying -1 on the RHS.
Thank you for the reply.. now I get what I did wrong...!
Can I ask you one more question??
The question I posted, my professor proved it by using triangle inequality like mathman did.
In the question, the promises are $|x-a|<\frac{r}{2}$ and $|y-b|<\frac{r}{2}$ and the conclusion we have to prove is $|(x+y)-(a+b)|. I thought if we have to prove something, we must start from the promises and end up with conclusion. But what my professor did and mathman did is they started from the part of the conclusion which is $|(x+y)-(a+b)|$ and end up with the sum of the two promises. is it possible to do it??

March 6th, 2013, 08:59 AM   #10
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Re: Inequality proof!!

Quote:
 Originally Posted by eChung00 I thought if we have to prove something, we must start from the promises and end up with conclusion.
You can also start with the RHS (or LHS, that depends however) of the statement of conclusion and then end up with the equivalence by the promises.

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