1. n*m=6+nm

Proof: Let n, m ∈ Z, such that n = 1 - 6/m. Then:

n * m = 6 + nm = 6 + (1 - 6/m)m = 6 + m - 6 = m.

m * n = 6 + mn = 6 + m(1 - 6/m) = 6 + m - 6 = m.

Thus, n is the identity of m. But n depends on m, and so, there is no single value n.

∴ No identity exists.

]]>The identity is a fixed number, e, such that:

a * e = e * a = a, for all a∈A (Where A in most of these cases is Z)

nm=6+nm

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.

n * m = n²m²

Same deal here. We want n²m² = m. Solve for n, you get it in terms of m.

n *m = min(n,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.

X * Y = X U Y

The null set is the identity.

n * m = n^m

No identity here because a *e = a, where e = 1, but e * a <> a.

So the majority of these have no identity. For the set one, show that a * e = e * a = a, where a is any set and e is the null set. It should be very striaghtforward.

However, the others are not. You need to show that for any element you pick, there exists another element such that the element you picked can't be the identity. I'll start working on one for you (I got some other things to do at the moment as well), but I should have it up in a few hours.

]]>It's false.

nm=(nm)²

so nm=0 or nm=1.

On Z+, n *m = min(n,m)

That's only if n=m and nn=n because otherwase nm>n and nm>m (n.m!=1)

so you get n=m=1

On Z+ nm=n^m

(n,1) and (2,2).]]>

a) On Z (integers), n * m = 6 + nm

b) On Z, n * m = n²m²

c) On Z+, n *m = min(n,m), the smaller of n and m

d) On P(A), for any set A, X * Y = X U Y

e) On Z+, n * m = n^m

I understand the concept of identities, i'm just unsure on how you can prove these things. I know if they don't have an identity, its easiest to show it by using a contradiction, but I'm having trouble finding any contradictions. Any help would be appreciated! Thanks.

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