Re: Interesting problem on symmetric difference

On 29 Jan 2007 23:29:23 +1100, Logan Lee <10464307@xxxxxxxxxx> wrote:


Let's assume also that you can apply distributive law to this.

You shouldn't just _assume_ things which have to be proved!

(It's _not obvious_ that the distributive law holds for the
delta-operation. Hence you have to prove this.)

well, distributive law was not given and i wasn't taught this
(as you say this is not obvious) but i've guessed it and it
occurs that at least for this case distributive law holds ;)

Hmmm...

A Delta (B Delta C) = (A Delta B) Delta (A Delta C) ???

Let A = B = C. Then we should have (if you were right):

A Delta (A Delta A) = (A Delta A) Delta (A Delta A)

for any set A. Let's see. We have:

(1) A Delta A = (A - A) + (A - A) = 0 u 0 = 0,
and
(2) A Delta 0 = (A - 0) + (0 - A) = A u 0 = A,

for any set A. And as a special case:

(3) 0 Delta 0 = 0.

Hence (1) (2)
A Delta (A Delta A) = A Delta 0 = A

and (1) (3)
(A Delta A) Delta (A Delta A) = 0 Delta 0 = 0.

Hence
A Delta (A Delta A) = (A Delta A) Delta (A Delta A)

would only hold for A = 0.

Hence in general:

A Delta (B Delta C) =/= (A Delta B) Delta (A Delta C).


You see: "Guessing is not good enough in/for math."


F.

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