MATH IS HARD

[Jasper]

Not plain old exponents.  I'm talking about negative exponents, in rational form, stuck inside fractions, and so on.  It wasn't that hard once I bothered to learn how, in retrospect.

Oh, and I keep forgetting my calculator.  That didn't help.

[Vene]

Not plain old exponents.  I'm talking about negative exponents, in rational form, stuck inside fractions, and so on.  It wasn't that hard once I bothered to learn how, in retrospect.

Oh, and I keep forgetting my calculator.  That didn't help.

Yes, the calculator is good.  Very, very good.

[Jasper]

TI-30X.

[BootyBay]

x=x

Clarify

(12 points)

Man, I'm a nerd for saying this, but here goes:

If x is a set:
Every member of x is a member of x.

If x is an element of a set:
if x = k, k a fixed element  implies k = x (because '=' is symmetric), implies x = x (because '=' is transitive),
which means, for any x equal to k, k is equal to x and x is equal to x.
if x does not equal k for any k belonging to any set, x is not an element of a set.

If x is a fixed element:
x is equal to itself.

[Jasper]

You, I like.

[rong]

x=x

Clarify

(12 points)

Man, I'm a nerd for saying this, but here goes:

If x is a set:
Every member of x is a member of x.

If x is an element of a set:
if x = k, k a fixed element  implies k = x (because '=' is symmetric), implies x = x (because '=' is transitive),
which means, for any x equal to k, k is equal to x and x is equal to x.
if x does not equal k for any k belonging to any set, x is not an element of a set.

If x is a fixed element:
x is equal to itself.

i believe it said "Clarify"

[Jasper]

And he did.  Or is it tl;dr for you?

[rong]

no, i just think it doesn't get any more clear than "x=x"

[Jasper]

There's a lot of room for misinterpretation in that statement, logically speaking.

[rong]

which statement?

[Jasper]

[rong]

 

[BootyBay]

I'm afraid I'm gonna have to go with rong on this one

[BootyBay]

More nerd points for anyone who can solve this (it's not a hw problem of mine; I'm far too advanced *gazes at self in mirror*)
For a topological space X,
X is Hausdorff iff the diagonal (the set {(x,x) | x in X}) is closed in (X, X)

[rong]

topology not my thing . . .


but . . .



why did the mathemetician name his dog Cauchy?





















cuz it left a residue at every pole!