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Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!

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snalin:
Is the images? not working for anyone else? If I open them in a new tab and replace "White" with "Black", it works. Huh.

ankhtahr:
'cause they're gifs and not pngs. That's why I recommended png.

snalin:
Fix'd!

Schmee:
This was in my uni's Maths and Stats society newsletter:

--- Quote ---Several years ago, A/Prof Barry Hughes set a first-year exam question which asked for a proof that there is no largest prime number.The following was sent to him by one of the graduate students who helped with the marking:

Hi Barry, just for educational value, I compiled a list of the most incorrect answers to question B4(b). It is entitled "15 good reasons why Pure Mathematics is not taught to first year students."

* Proof by example: “Let x be the largest prime. Then x = 91 but 91 + 6 = 97, which is prime. Therefore, 91 cannot be the largest prime number. Therefore there is no largest prime number.”
* Proof by oddness: “If n is the largest prime number, then n is odd. Then (n + 1)=2 is even. Therefore, (n + 1)=2 + n is odd. But (n + 1)=2 + n is not divisible by any number except itself. As it is bigger than n, the assumption is wrong, by contradiction.”
* Proof by intuition: “Prime numbers are integers that can be divided by themselves only; prime numbers are odd with the exception of 2. By intuition as n -> infinity, there will always be an odd number that cannot be divided by any other number besides itself.”
* Proof by sqrt(2): “that p is divisible by q, i.e. p/q = 2r, where r is an even number. Then p = 2qr so p2 = 4q2r2. But r2 does not exist and q! = 1. Therefore, q must exist. Since q exists, p must be divisible. Therefore, by contrary, there is no largest prime number.”
* Proof by superinduction: “2 is a prime number. Now assume N is the largest prime. But then N + 1 exists and is also prime. Therefore, by induction there is no highest prime number.
* Proof by the previous question: “Suppose N is the largest prime. Then let N = (n2)/2. Therefore n = sqrt(2N). But from above (1 + 2 + 3 + ... +n) > N. Hence, there is a larger prime number than N.”
* Proof by tutorial question: “Let m, n be two integers with m > n + 1. If k is even, mk + nk cannot be expressed in terms of (m + n)(polynomial in m and n) and so is prime. Therefore, as m and n can be any numbers, there is obviously no largest prime number.”
* Proof by having no idea what a prime is: "Say the largest prime possible is x, then 2x is also a prime since the statement is true for all natural numbers."
* Proof by experimental data: "Suppose n is the highest prime. Then 2n-1 is also prime. But 2n-1>n so there is no highest prime (Check: 2*2-1=3, 2*3-1=5, 2*5-1=11, 2*11-1=23, so true)"
* Proof by subscript: “If there is a highest prime, we can number all the primes p1, p2, ... , pn. But as there is no highest natural number, there is always an n + 1 so there must be a pn+1. Therefore, there is no highest prime.”
* Proof by infinity: “Let n be the highest prime number. But infinity is greater than all numbers so infinity > n. If n is the highest prime this would mean infinity has factors. Therefore, we have a contradiction.”
* Proof by reverse logic: “All prime numbers are odd. Suppose there were a highest prime. Then we have a highest odd number. But if 2k+1 is the highest odd number, then 2k+3 = 2(k+1)+1 is also an odd number. Therefore, we have a contradiction and therefore we have a contradiction."
* Proof by denial: “Assume there is a largest prime M. We can add 1 to M until we get another prime number N (M + 1 + 1 + 1 + ... + 1 = N). But then N > M. Therefore, M is not the largest prime number, so there is no largest prime number.”
* Proof by formula: “As prime numbers are derived via the formula, we can assume it works for n = k giving the highest prime number. But then it also works for n = k + 1, so there is no highest prime number.”
* Proof by continuity: “Let x be the largest prime number. Then x > all other primes. But then (x + n), the next prime number, does not exist. However, numbers are continuous and so (x + n) does exist. Therefore, there is no x.”These are all verbatim answers from the 150-odd papers I marked, i.e. about one in ten. My favourite is definitely number 10; I think that is quite ingenious.
— George Doukas

--- End quote ---
... That took a while to copy over, but I think you'll agree it was worth it.

Barmymoo:
What would be a correct answer to that question?

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