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Do all existence proofs proceed indirectly (a.k.a "by contradiction," "reductio ad absurdum," etc.)?

For example, can we prove directly that in a field the additive identity element and the multiplicative identity element are distinct? I have not found any proof for this that proceeds directly.

So I'm looking for some similar existence lemma that might provide hints by analogy?

If I'm on a fool's errand (direct proof is impossible), please explain or prove this.

Thank you.

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There is only one ring in which the additive and multiplicative identity are equal, namely the zero ring, which is explicitly excluded from the definition of a field. –  Dustan Levenstein Jul 15 at 20:14
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Short answer to the title: No. –  Aryabhata Jul 15 at 21:37

2 Answers 2

From the other field axioms (commutivity, associtivity, etc) you cannot prove that the additive and multiplicative identity are distinct. That is why it is included (although sometimes overlooked) as an axiom of fields.

To show that you cannot prove that $0_{\text{Field}} \ne 1_{\text{Field}}$ all you have to do is give a structure that satisfies all of the field axioms except $0 = 1$. An example (the only one) is the domain with only 1 element, and addition and multiplication are defined as identity functions. It's a simple enough excerise to show that such a structure satisfies all field axioms except $0 \ne 1$.

Btw, you can also replace $0 \ne 1$ among field axioms with "the domain of the field contains at least 2 values".

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The standard proof for the uniqueness is a direct one.

Let $a$ be an additive identity, and $b$ be an identity. Since $a = a + b = b$, $a = b$.

Note that we say nothing about whether $a$ and $b$ are distinct, and so there is no assumption for us to contradict. We simply derive that they are equal, and that's all we wanted. It's the same with the multiplicative identity.

EDIT: It just occurred to me, you might be asking a different question, whether $0 \ne 1$. That is usually accepted as part of the definition of a field.

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