Question on a step in my proof.

[Diffy]

I don;t think the actual problem I am trying to prove is relevant. But I have got to a step in my proof where I must show that | a - b | = | -(a - b)|

So by definition
|a - b| = +/- (a-b)

and |-(a - b)| = +/- (- (a - b)) = +/- (a-b)

Thus the two are equal.

Is this acceptable?

Or do I really need to break it down into cases where a - b is Positive negative or 0?

Thanks.

 

[Dick]

|a-b|=+/-(a-b) is NOT the definition of absolute value. The definition of |a-b| is that it is (a-b) if a-b>=0 and -(a-b) if a-b<0. You do need to break it into cases. But it's easier to prove |x|=|-x| (by cases) and then set x=a-b.

 

[Diffy]

Got it, yes, I realize I can say |a-b| is greater than or equal to +/- (a-b).

but certainly not equal.

I showed that |x| = |-x| as follows:

case x>0:
If x > 0 then |x| = x by definition. If x>0 then -x<0 since x and -x can't both be in R+. if -x < 0 then |-x| = -(-x) = x

case x = 0. If x = 0 then -x = 0 and |x| = |0| = 0 and |-x| = |0| = 0.

case x < 0. If x<0 then |x| = -x by definition. If x<0 then -x>0 since x and -x can both be in R+. since -x>0, |-x| = -x and we are done.

 

[Dick]

Got it, yes, I realize I can say |a-b| is greater than or equal to +/- (a-b).

but certainly not equal.

I showed that |x| = |-x| as follows:

case x>0:
If x > 0 then |x| = x by definition. If x>0 then -x<0 since x and -x can't both be in R+. if -x < 0 then |-x| = -(-x) = x

case x = 0. If x = 0 then -x = 0 and |x| = |0| = 0 and |-x| = |0| = 0.

case x < 0. If x<0 then |x| = -x by definition. If x<0 then -x>0 since x and -x can both be in R+. since -x>0, |-x| = -x and we are done.

That proof is just fine. If you know |ab|=|a||b|, you can also prove it by |-x|=|(-1)*x|=|-1|*|x|=|x|.

 

[tiny-tim]

… I have got to a step in my proof where I must show that | a - b | = | -(a - b)|

So by definition
|a - b| = +/- (a-b)

and |-(a - b)| = +/- (- (a - b)) = +/- (a-b)

Thus the two are equal.

Is this acceptable?

Hi Diffy!

If your main proof required you to show that |a - b| = |b - a|,

then I would say that it is enough to say (a - b) = -(b - a), and therefore |a - b| = |b - a|.

But to show that | a - b | = | -(a - b)|, in that form, doesn't ssem to me to require any proof … |x| = |-x| is part of the definition of ||.

(btw, using the symbol +/-, or even ±, would not be acceptable … you should use something more rigorous, like √(x²), or sup{x,-x}, or write the concept out in English )

 

[Dick]

If x<0 then -x>0 since x and -x can both be in R+.

That sounds a little funny. If a<b then -a>-b. That's why x<0 implies -x>0.

 

[Diffy]

Dick, there was a mistake,

It should say if x<0 then -x>0 since x and -x can't both be in R+.

I don't have that if a<b then -a>-b anywhere, but I do have in the my order property that only one of the following is true:
a in R+
a = 0
-a in R+

I was attempting to use that. Thanks a lot for your help dick!

Hey TinyTim,

It wasn't immediately clear (at least to me) from my definition so I felt the need to prove it.
I had:
|a| = a if a>0
|a| = 0 if a = 0
|a| = -a if a < 0

Thanks for your help Tiny-Tim!

 

[Dick]

Dick, there was a mistake,

It should say if x<0 then -x>0 since x and -x can't both be in R+.

I don't have that if a<b then -a>-b anywhere, but I do have in the my order property that only one of the following is true:
a in R+
a = 0
-a in R+

I was attempting to use that. Thanks a lot for your help dick!

Ok. That makes sense.

 

[tiny-tim]

Hey TinyTim,

It wasn't immediately clear (at least to me) from my definition so I felt the need to prove it.
I had:
|a| = a if a>0
|a| = 0 if a = 0
|a| = -a if a < 0

Yes, that method's fine!