Fun Stuff > CHATTER
Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Carl-E:
Euclid had a good one. If there's a largest prime, then you list them all and multiply all of the primes together into one big number.
Then add 1 to that result. It's bigger than the biggest prime, so it's not prime. And none of the primes will divide it, they'll all have a remainder of 1. That means that it must be prime, since nothing divides it but 1 and itself.
And that's absurd - it can't be both prime and not prime, so the assumption (that there's a largest prime) is false, there isn't one.
Ankhtahr, did you get the 52n - 2n yet? It's (52)n - 2n, which will always factor as
(52 - 2)((52)n-1 + (52)n-22 + ... + 2n-1), and so there's always a factor of 23.
(sometimes, it just looks like it needs induction...)
ankhtahr:
Schmee: The first one cracked me up already. Where the hell did that 91 come from?
"Proof by having no idea what a prime is"
Bwahaha!
Carl: Yes, I had to turn in the exercise sheet on Thursday already.
Oh, and I'm so very sorry to say, but Snalin: I found a much, much simpler proof.
Like I had said, I suspected the whole thing could be solved with the binomial theorem (). I noticed that you can just choose a = 1, and apply it on the term.
Which makes it look like this:
Carl-E:
"Therefore, we have a contradiction, and so therefore, we have a contradiction"
Pure gold!
Loki:
--- Quote ---Prove: Each convergent series over K \in {\mathds{R}, \mathds{C}} is bounded.
--- End quote ---
Is there any sensible answer except "well. duh, if it converges to x, it is bounded by x"?
Carl-E:
--- Quote from: Loki on 10 Nov 2013, 07:42 ---
--- Quote ---Prove: Each convergent series over K \in {\mathds{R}, \mathds{C}} is bounded.
--- End quote ---
Is there any sensible answer except "well. duh, if it converges to x, it is bounded by x"?
--- End quote ---
False!
Just in R, the sequence 1, -1, 1/2, -1/2, 1/3, -1/3, ... converges to 0, but is bounded above by 1 and below by -1. The limit is rarely a bound unless the sequence is monotone.
Though maybe I don't understand the question - what exactly is {\mathds{R}, \mathds{C}}? It looks like the two-dimensional space R X C, which doesn't really change the answer, but I don't think that's what it means?
(Topology is almost nothing but a collection of counterexamples... I got million of 'em!)
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