Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

Login

Username

Password

Not registered yet?

Post a reply

Go back

Write your message and submit
:) :| :( :D :o ;) :/ :P :lol: :mad: :rolleyes: :cool: | :dizzy :eek :kiss :roflol :rolleyes :shame :down :up :touched :sleep :wave :swear :tongue :what :faint :dunno
Options

Go back

Topic review (newest first)

Ricky
2006-03-09 03:11:43

I'll do the first as an example, you try to do the rest:

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.

Ricky
2006-03-09 03:05:14

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.

Ricky
2006-03-09 02:09:40

krassi, I don't think you're trying to find the right thing.

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 = nm

Same deal here.  We want nm = 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.

krassi_holmz
2006-03-08 17:33:21

nm=6+nm?
It's false.
nm=(nm)
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).

Kazy
2006-03-08 14:58:22

I need to determine which of the following have identities with proofs.
a) On Z (integers), n * m = 6 + nm
b) On Z, n * m = nm
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.

Board footer

Powered by FluxBB