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#1 2006-03-26 07:06:53

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

Proving rational inverses

I need to prove that every element of Q (rational numbers) has an inverse with respect to addition. From what understand, I need to use the fact that [0,1] is the additive identity of Q. However, I have no idea what to do beyond that.

Also, I need to prove that every element of Q except [0,1] has an inverse with respect to multiplication. Again, to do so, I'm thinking it has something to do with the fact that [1,1] is the miltiplicative identity of Q.

If anyone can help, that'd be great!

Last edited by Kazy (2006-03-26 07:07:45)

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#2 2006-03-26 08:35:09

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Proving rational inverses

I need to prove that every element of Q (rational numbers) has an inverse with respect to addition. From what understand, I need to use the fact that [0,1] is the additive identity of Q.

0 is the additive identity of Q.  And you can have 1 and only 1 additive identity.

As for a proof, I would just say let a/b be integers such that b is nonzero.  Then a/b + -a/b = 0, and thus, every rational number has an additive inverse.

Also, I need to prove that every element of Q except [0,1] has an inverse with respect to multiplication. Again, to do so, I'm thinking it has something to do with the fact that [1,1] is the miltiplicative identity of Q.

[1,1] is better written as just 1.  The only rational number that doesn't have a multiplicative inverse is 0.  Consider a/b where a and b are integers and a and b are nonzero.  Then a/b * b/a = 1, and thus, all nonzero rationals have a multiplicative inverse.

Last edited by Ricky (2006-03-26 08:36:04)


"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..."

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