Absolute Value Proof

[mathman123]

When testing this out for a bunch of numbers, it appears true, but I was wondering if anyone had a proof for it:

abs(a-b) is greater than or equal to abs(abs(a)-abs(b))

for all a,b

 

[tkhunny]

Consider if a > b.
Consider if a < b.
Consider if a = b.

See if that leads anywhere useful.

 

[burakaltr]

When testing this out for a bunch of numbers, it appears true, but I was wondering if anyone had a proof for it:

abs(a-b) is greater than or equal to abs(abs(a)-abs(b))

for all a,b

Since both left of >= and right are greater than zero, take square of both sides

you get :

a**2 - 2 a b + b**2>=a**2+b**2 - 2 abs(a)* abs(b)

Shift positions ( left to right and right to left )

2ab<= 2 abs(a)*abs(b)

yields

ab<= abs(a)*abs(b)

OR

ab<= abs(a*b)

this always holds true.

 

[Deleted member 4993]

when testing this out for a bunch of numbers, it appears true, but i was wondering if anyone had a proof for it:

abs(a-b) is greater than or equal to abs(abs(a)-abs(b))

for all a,b​

since both left of >= and right are greater than zero, take square of both sides

you get :

A**2 - 2 a b + b**2>=a**2+b**2 - 2 abs(a)* abs(b)

shift positions ( left to right and right to left )

2ab<= 2 abs(a)*abs(b)

yields

ab<= abs(a)*abs(b)

or

ab<= abs(a*b) <<< this is not what the original poster wanted

this always holds true.

.​

 

[burakaltr]

.[/indent]

You do NOT get my point.

 

[pka]

When testing this out for a bunch of numbers, it appears true, but I was wondering if anyone had a proof for it:
abs(a-b) is greater than or equal to abs(abs(a)-abs(b)) for all a,b

This proof depends upon this:
\(\displaystyle -|x|\le y\le |x|\text{ implies }|y|\le |x|.\)

Proof
\(\displaystyle |a|\le |a-b|+|b|\text{ implies }|a|-|b|\le |a-b|\)

Likewise \(\displaystyle |b|-|a|\le |b-a|=|a-b|.\)

Thus \(\displaystyle -|a-b|\le |a|-|b|\le |a-b|\) so \(\displaystyle ||a|-|b||\le |a-b|\)

 

[Deleted member 4993]

Since both left of >= and right are greater than zero, take square of both sides

you get :

a**2 - 2 a b + b**2>=a**2+b**2 - 2 abs(a)* abs(b)

Shift positions ( left to right and right to left )

2ab<= 2 abs(a)*abs(b)

yields

ab<= abs(a)*abs(b)

OR

ab<= abs(a*b)

this always holds true.

You do NOT get my point.

I do.

You have proven "something correct" by assuming that the proposition to be proven is true.

Suppose we assumed ** means addition.

Then you said 2**2 = 2 + 2 = 4

Hence ** means addition.

That is of course not correct.

You are trying to use inverse of "reductio ad absurdum" - it does not work.

 

[burakaltr]

I do.

You have proven "something correct" by assuming that the proposition to be proven is true.

Suppose we assumed ** means addition.

Then you said 2**2 = 2 + 2 = 4

Hence ** means addition.

That is of course not correct.

You are trying to use inverse of "reductio ad absurdum" - it does not work.

But Sir,

** means to the power of.

I was using FORTRAN notation

 

[Deleted member 4993]

I also use FORTRAN - so I know what you mean by **.

Lets rephrase my proposition:

Suppose I claim: → "Power" (**) means same as "addition"

Then I offer proof:

2**2 = 4

2+2 = 4 (true)

Hence

** = +

But proof above is not correct.

Your proof-logic is similar to this logic.

 

[burakaltr]

I also use FORTRAN - so I know what you mean by **.

Lets rephrase my proposition:

Suppose I claim: → "Power" (**) means same as "addition"

Then I offer proof:

2**2 = 4

2+2 = 4 (true)

Hence

** = +

But proof above is not correct.

Your proof-logic is similar to this logic.

No Sir,

My Proof is complete and I think we can ask other members on its correctness.

What do you maths people think ?

Please commentate further .

Thanks.

 

[Mrspi]

I

No Sir,

My Proof is complete and I think we can ask other members on its correctness.

What do you maths people think ?

Please commentate further .

Thanks.

I agree with Subhotosh Khan....you BEGAN your proof by assuming the statement you were trying to prove. You started with the proposition you were supposedly proving, and then said "square both sides." And proceeded from there.

Clearly that is not the same thing as starting with something KNOWN to be true, and by using legitimate algebraic processes, arriving at the the statement you're attempting to prove true.

 

[burakaltr]

I

I agree with Subhotosh Khan....you BEGAN your proof by assuming the statement you were trying to prove. You started with the proposition you were supposedly proving, and then said "square both sides." And proceeded from there.

Clearly that is not the same thing as starting with something KNOWN to be true, and by using legitimate algebraic processes, arriving at the the statement you're attempting to prove true.

Thanks for your answer.

And I would like to to pose this question : If I rewind ( go backwards from finish to the beginning ) my proof, what would have i done wrong ?

Thank You.

 

[pka]

And I would like to to pose this question : If I rewind ( go backwards from finish to the beginning ) my proof, what would have i done wrong ?

Let’s be very clear on what makes an acceptable proof.
Start with a known fact.
\(\displaystyle {-2ab \ge -2|ab|}\)
\(\displaystyle a^2-2ab +b^2 \ge |a|^2-2|ab| + |b|^2\)
\(\displaystyle (a-b)^2 &\ge (|a|-|b|)^2\)
\(\displaystyle |a-b| \ge |\,|a|-|b|\,| \)

The last inequality comes from this fact:
\(\displaystyle K^2\le J^2\text{ if and only if }|K|\le |J|.\)

 

[lookagain]

Proof by contradiction
----------------------

\(\displaystyle Assume \ |a - b| \ < \ \bigg| |a| - |b|\bigg|\)


is equivalent to


\(\displaystyle \sqrt{(a - b)^2} \ < \ \sqrt{(|a| - |b|)^2}\)


because \(\displaystyle |x| \ = \ \sqrt{x^2}\)

Each side is positive. Square each side:


\(\displaystyle (a - b)^2 \ < \ (|a| - |b|)^2\)


\(\displaystyle a^2 - 2ab + b^2 \ < \ |a|^2 - 2(|a|)(|b|) + |b|^2\)


\(\displaystyle Note: \ \ (|a|)(|b|) \ = \ |ab|\)


\(\displaystyle a^2 - 2ab + b^2 \ < \ a^2 - 2|ab| + b^2\)


\(\displaystyle -2ab \ < \ -2|ab|\)


\(\displaystyle \frac{-2ab}{-2} \ > \ \frac{-2|ab|}{-2}\)


\(\displaystyle ab \ > \ |ab|\)


This is false, because the left-hand side can equal a negative number,
while the right-hand side is always nonnegative. This is a
contradiction.


Then the assumption made by me is false, so the statement
of the original problem must be true.