[John JD Dorian]

If anyone has an interesting math problem to talk about, or needs help with something they're working on, bring it in here!

[John JD Dorian]

no one, huh...

[Fullblast]

Do you know anything about Boolean Logic? Stuff makes my head hurt....

[John JD Dorian]

Originally Posted by Fullblast

Do you know anything about Boolean Logic? Stuff makes my head hurt....

i very much do. ask away.

[Truman122]

Do mathematical physics problems count...?

[John JD Dorian]

Originally Posted by Truman122

Do mathematical physics problems count...?

no reason not to give it a try.

[perceptrons]

I can also help out with any boolean logic questions.

[John JD Dorian]

someone ask something!!!

[John JD Dorian]

or i will...

[tonyC4L]

This question was on my analysis midterm last week:

Let f: R -> R be defined by f(x) = 0 if x is rational, x if x is irrational. Show that lim x->0 f(x) = 0 but lim x->x0 f(x) does not exist for x0 != 0.

Epsilon-delta proofs are welcome along with equivalent arguments using sequences.

[John JD Dorian]

Originally Posted by tonyC4L

This question was on my analysis midterm last week:

Let f: R -> R be defined by f(x) = 0 if x is rational, x if x is irrational. Show that lim x->0 f(x) = 0 but lim x->x0 f(x) does not exist for x0 != 0.

Epsilon-delta proofs are welcome along with equivalent arguments using sequences.

what you are essentially proving is that f is continuous at 0 but nowhere else. alright, here goes:

Proof:
(i) At x=0, f is continuous. For if we let e > 0, we can choose d = e, and then if |x - 0| = |x| < d, then |f(x) - f(0)| = |f(x)| is either 0 for x in Q or |x| for x not in Q. In the former case, we have |f(x)| = 0 < e, and in the latter we have |f(x)| = |x| < d = e, so we are done.

(ii) Let x ≠ 0. Then, f is not continuous at x. To show this, we choose e = |x| > 0, and let d > 0.

If x is in Q and x > 0, then we are sure that there is some y in the interval (x, x+d) such that y is not in Q, so that f(y) = y. If x < 0, we can similarly choose a y in the interval (x-d, x). Either way, for this y we have |x-y| < d, yet |f(x) - f(y)| = |0 - f(y)| = |f(y)| = |y| > |x| = e, so f is not continuous at x.

If x is not in Q, then we are sure that there is some y in the interval (x-d, x+d) such that y is in Q, so that f(y) = 0. But for this y, we have |x-y| < d, yet |f(x) - f(y)| = |f(x)| = |x| = e, so f is not continuous at x.

In either case, f is not continuous at x.

////

[BPerone201]

I'm probably going to be popping up in this forum a lot.
I've been failing Algebra 2 since last year and repeating.

[John JD Dorian]

Originally Posted by BPerone201

I'm probably going to be popping up in this forum a lot.
I've been failing Algebra 2 since last year and repeating.

please, bring it on!

[tonyC4L]

Originally Posted by John JD Dorian

what you are essentially proving is that f is continuous at 0 but nowhere else. alright, here goes:

Proof:
(i) At x=0, f is continuous. For if we let e > 0, we can choose d = e, and then if |x - 0| = |x| < d, then |f(x) - f(0)| = |f(x)| is either 0 for x in Q or |x| for x not in Q. In the former case, we have |f(x)| = 0 < e, and in the latter we have |f(x)| = |x| < d = e, so we are done.

(ii) Let x ≠ 0. Then, f is not continuous at x. To show this, we choose e = |x| > 0, and let d > 0.

If x is in Q and x > 0, then we are sure that there is some y in the interval (x, x+d) such that y is not in Q, so that f(y) = y. If x < 0, we can similarly choose a y in the interval (x-d, x). Either way, for this y we have |x-y| < d, yet |f(x) - f(y)| = |0 - f(y)| = |f(y)| = |y| > |x| = e, so f is not continuous at x.

If x is not in Q, then we are sure that there is some y in the interval (x-d, x+d) such that y is in Q, so that f(y) = 0. But for this y, we have |x-y| < d, yet |f(x) - f(y)| = |f(x)| = |x| = e, so f is not continuous at x.

In either case, f is not continuous at x.

////

Very nice, thank you. We actually have not learned the definition of continuity yet in my class, so I didn't make that connection when taking the test, but after reading the definition and then your proof it all makes sense.

Originally Posted by BPerone201

I'm probably going to be popping up in this forum a lot.
I've been failing Algebra 2 since last year and repeating.

well, the thing about us math people is that we like doing math. so please, ask questions!

[tonyC4L]

And for the record I'm mostly interested in problems in differential equations (ODEs and PDEs). I'm also taking complex analysis right now and I like that too.