Math Is Fun Forum

Dilbert

Member

Registered: 2006-09-05

Posts: 12

Hey

How would you show that a binary operation is closed, when dealing with an algebraic example like the following?
a*b = a+b - ab

hmmm?

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Hey

Well, first you have to decide what set you are working with.  If that set is closed under addition, subtraction, and multiplication, then you can know it's also closed on a+b-ab.

If the set is finite, then just try every possibility.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Dilbert

Member

Registered: 2006-09-05

Posts: 12

Re: Hey

But I'm dealing with letters, not numbers, you cannot operate on letters. It says defined on R. This is abstract.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Hey

Letters are variables.  More specifically, variables in R.

Since R is closed under addition, multiplication, and subtraction, it must be closed under a + b - ab.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Dilbert

Member

Registered: 2006-09-05

Posts: 12

Re: Hey

But how would I prove it?

Dross

Member

Registered: 2006-08-24

Posts: 325

Re: Hey

Simply state that, since R is closed under multiplication, ab is in R. Then, since R is closed under addition (and subtraction), (b - ab) is in R and a + (b - ab) is in R. Thus R is closed under said operation.

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Bad speling makes me [sic]

Dilbert

Member

Registered: 2006-09-05

Posts: 12

Re: Hey

Ah great! thanks

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Hey

Dross's proof is absolutely correct.  But normally such statements are uncessary.  Simply stating that R is close under addition, subtraction, and multiplication is enough to prove that a + b - ab is closed.

For example, when you do more advanced proofs, you can simply state that an even times an even is divisble by 4, there is no need to prove it.

But then again, it also depends on your teacher/professor, and what type of course it is.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Dilbert

Member

Registered: 2006-09-05

Posts: 12

Re: Hey

Much appreciated Ricky, I know I've alot to learn, i'm just trying my best.

Dilbert

Member

Registered: 2006-09-05

Posts: 12

Re: Hey

Back again, same question, need to prove it is associative. My answer:

Q. a*b = a+b-ab

A. If associative: (a*b)*c = a*(b*c)

a*(b*d) = a +(b*d) - a(b*d)
So: ab + ad = a +bd - (ab+ad)

(a*d)*b = (a*b) + d - ((a*b) *d)
So: ad + db = ab + d - (ad + bd)

hmmm?

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Hey

Except for a minor typo which through me off, that is correct.

So is it that case that a +bd - (ab+ad) = ab + d - (ad + bd)?  The answer to this is the answer to whether or not * is associative.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Dilbert

Member

Registered: 2006-09-05

Posts: 12

Re: Hey

a +bd - (ab+ad) = ab + d - (ad + bd)
Let a = 1, b=2, d=3

7-5 = 5-9
2 = -4

So it's not associative?

Hmm, but the question was "show that (said operation) is associative" ???

Last edited by Dilbert (2006-09-06 12:44:45)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Hey

Whoops, we both made the same error.

Don't confuse ab with a*b.  They are entirely different.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Dross

Member

Registered: 2006-08-24

Posts: 325

Re: Hey

Ah yes, I was veeeery confused by what you wrote earlier, Dilbert!

You can show a*b = a + b - ab to be associative in the following way:

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Bad speling makes me [sic]

Dilbert

Member

Registered: 2006-09-05

Posts: 12

Re: Hey

That looks right, but some of the steps are confusing.

Dross

Member

Registered: 2006-08-24

Posts: 325

Re: Hey

Which ones are confusing? What's happening may be more obvious if you go from the end and work towards the middle.

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Bad speling makes me [sic]

Dilbert

Member

Registered: 2006-09-05

Posts: 12

Re: Hey

I was getting confused on a few steps. The * sign is confusing, you think of multiplication. It's subtle, you have to think forwards, then backwards.
But now I understand, well done.

krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: Hey

If this operation is confusing, you can replace i6t by a function, for example:

Then you should prove that :

Now, simplifying the two sides:
1.

2.

Another observation: because (a+b) and (-ab) are symmetric, then a*b is commutative, e.a. a*b=b*a.

Last edited by krassi_holmz (2006-09-07 04:52:44)

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Hey

Or simply replace it with a different symbol.  # is fairly common.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: Hey

Another common is circled plus or circled dot or circled star:

Last edited by krassi_holmz (2006-09-08 01:23:14)

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IPBLE:  Increasing Performance By Lowering Expectations.