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

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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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 22,030

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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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 22,030

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

Thanks!

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