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## #1 2006-03-23 16:41:30

Kazy
Member
Registered: 2006-01-24
Posts: 37

### Rational Numbers

I need to prove the following:

a) Addition on Q is communitive.
b) Multiplication on Q is associative.
c) Multiplication on Q is communitive.

I figured out how to prove addition on Q is associative.. Here is what I did:

([a,b] + [c,d]) + [e,f]
= [ad + bc, bd] + [e,f] = [(ad+bc)f + (bd)e, (bd)f]
= [adf+bcf+bde, bdf] = [a(df) + b(cf+de), b(df)]
= [a,b] + [cf + de, df] = [a,b] + ([c,d] + [e,f])

I can't figure the others out. any help would be appreciated.

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## #2 2006-03-23 17:28:02

ganesh
Administrator
Registered: 2005-06-28
Posts: 23,368

### Re: Rational Numbers

Addition on Q is commutative:-
[a,b,c,d belong to integers, b and d are not equal to zero]
a/b + c/d = (ad+bc)/bd
c/d + a/b = (bc+ad)/bd = ((ad+bc)/bd
From the above,
a/b + c/d = c/d + a/b.
Therefore, addition on Q is commutative.

a/b x c/d = ac/bd
c/d x a/b = ca/db = ac/bd

From the above,
a/b x c/d = c/d x a/b
Therefore, multiplication on Q is commutative.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #3 2006-03-23 18:14:50

ganesh
Administrator
Registered: 2005-06-28
Posts: 23,368

### Re: Rational Numbers

Oh..I didn't notice....you had asked for associativity on Q too...

Let a,b,c,d,e,f be integers, b,d,e not equal to zero.

a/b x (c/d x e/f) = a/b x ce/df = ace/bdf

(a/b x c/d) x e/f = ac/bd x e/f = ace/bdf

Therefore, multiplication on Q is associiative.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #4 2006-03-24 05:51:23

Kazy
Member
Registered: 2006-01-24
Posts: 37

Thanks!

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