Unique Identities: Proving O₁ = O₂ in Theorem for Proof of Identity

  • #1
ver_mathstats
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Homework Statement
Prove the identity of plus.
Relevant Equations
Declaration: O₁, O₂ : ℤ

Axiom “Left-identity of +”: x = O₁ + x
Axiom “Right-identity of +”: x = x + O₂

Now, prove the following.
Theorem “Identities of + are unique”: O₁ = O₂
Theorem “Identities of + are unique”: O₁ = O₂
Proof:
O₁
= Left Identity of +
O₁ + x

I'm a little confused where to begin this proof, I don't know if that is the first step either I think it is. Proofs are not a strength of mine so I struggle to see how to show that O₁ = O₂. Any guidance would be appreciated, thank you.
 
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  • #2
Take your two defining equations and set ##x=0_1## in one and ##x=0_2## in the other.
 
  • #3
ver_mathstats said:
Homework Statement:: Prove the identity of plus.
Relevant Equations:: Declaration: O₁, O₂ : ℤ

Axiom “Left-identity of +”: x = O₁ + x
Axiom “Right-identity of +”: x = x + O₂

Now, prove the following.
Theorem “Identities of + are unique”: O₁ = O₂

Left Identity of +
At first glance, I didn't know what you were trying to do, but maybe it's a language translation thing.

A more usual phrasing would be "O is the additive identity" or "O1 is the left-additive identity".
My point is that O1 and O2 are the left/right addition identities, the things that you can add to a number without changing it.

In a similar vein there is the concept of a multiplicative identity.
 
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  • #4
Mark44 said:
At first glance, I didn't know what you were trying to do, but maybe it's a language translation thing.
One can define only one sided neutrals in group theory and show that they have to be the same, even in non commutative groups. I just don't recall whether one one sided neutral is already sufficient, or whether both are needed.
 
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