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 November 24th, 2018, 09:21 AM #1 Senior Member   Joined: Nov 2011 Posts: 250 Thanks: 3 The unique multiplicative inverse How can prove that the multiplicative inverse is unqiue? November 24th, 2018, 09:32 AM   #2
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 Originally Posted by shaharhada How can prove that the multiplicative inverse is unqiue?
If a is invertible and 1 is the multiplicative identity, then

$1 = ab = ac$ implies $b = c$ by multiplying by $a^{-1}$ on the left. November 24th, 2018, 09:35 AM #3 Senior Member   Joined: Nov 2011 Posts: 250 Thanks: 3 two... Why can't be 2 or 3 multiplactive identities? November 24th, 2018, 09:40 AM   #4
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Quote:
 Originally Posted by shaharhada Why can't be 2 or 3 multiplactive identities?
Did you read post #2? You thanked it. November 24th, 2018, 09:54 AM   #5
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 Originally Posted by shaharhada Why can't be 2 or 3 multiplactive identities?
1 isn't the usual 1. It's the symbol for the multiplicative identity (in a group, say). So if 2 were the multiplicative identity in some group, then 2 would be denoted as 1. Subtle point.

But in general if $ab = ac$ and $a$ is invertible, then $b = c$ for the same reason as above. I don't even need the symbol 1 for this proof. November 24th, 2018, 10:05 AM #6 Senior Member   Joined: Nov 2011 Posts: 250 Thanks: 3 What will be if there is more than 1 such as two of or three of the multiplicative identity? What will happen then? What there is only and only one multiplicative identity? November 24th, 2018, 10:19 AM   #7
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 Originally Posted by Maschke If a is invertible and 1 is the multiplicative identity, then $1 = ab = ac$ implies $b = c$ by multiplying by $a^{-1}$ on the left.
While this is certainly a correct answer to the question the OP asked, I think the OP actually means to ask how we know the multiplicative identity is unique. If so then your proof, which assumes this is true, probably explains his/her confusion.

@OP: If you are actually trying to prove that the multiplicative identity is unique, then you should note that the invertible elements of a ring form a group under multiplication. So the identity is unique provided that the identity in a group is unique. This is essentially a 1 line proof which you should do as an exercise.

Specifically, suppose $(G,*)$ is a group and $1, 1'$ are both identity elements. Prove that $1 = 1'$. November 25th, 2018, 08:33 AM #8 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2643 Math Focus: Mainly analysis and algebra I presume that the said proof is something like if $1$ is an identity element then $1 * 1' = 1'$ and if $1'$ is an identity element then $1 * 1' = 1$ so if both are identity elements we have $$1 = 1 * 1' = 1'$$ Thanks from SDK November 25th, 2018, 04:14 PM   #9
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 Originally Posted by v8archie I presume that the said proof is something like if $1$ is an identity element then $1 * 1' = 1'$ and if $1'$ is an identity element then $1 * 1' = 1$ so if both are identity elements we have $$1 = 1 * 1' = 1'$$
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