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I'll do the first as an example, you try to do the rest:
Actually, nevermind. Just solve for m like I did (for most of them), and show that this "equation" is the identity, but this equation also is not a single number.
krassi, I don't think you're trying to find the right thing.
We wish to find an n such that 6+nm = m. So solving for n, we get n = 1 - 6/m. But n depends on m. Thus, there exists no single n such that 6+nm = m, so no identity exists.
Same deal here. We want n²m² = m. Solve for n, you get it in terms of m.
Again, does not exist. Pick any number in Z+, and you can always find a number greater than it (Archimedian Principle). Thus, min(n, m) will never always be m for any fixed n.
The null set is the identity.
No identity here because a *e = a, where e = 1, but e * a <> a.
I need to determine which of the following have identities with proofs.