## [0882] Bonus Question

└ posted on Monday, 1 May 2017, by Novil

I wrote a blog post with more info about the *Sandra and Woo* adventure game!

**Sheet:** Bonus question (5 points)

Christian has discovered an intriguing property of prime numbers. He immediately tells his best friend Leonhard about it. But Leonhard is skeptical about Christian’s claim:

“Every even number greater than 2 can be expressed as the sum of two prime numbers!”

Examples:

4 = 2 + 2

6 = 3 + 3

8 = 5 + 3

10 = 5 + 5 or 7 + 3

**Sandra:** … Kannst du Christians Behauptung beweisen oder widerlegen?

**Sandra:** Mrs. Gallagher?
**Mrs. Gallagher:** Yes…?

**Sandra:** That should be worth at least 10 bonus points!
**Mrs. Gallagher:** I’ll think about it.

You can’t, it’s just the way numbers work

Question: Can you prove or disprove Christian’s claim?

Answer: No.

@ Titan:That’s definitely a novel way of answering the problem that has stumped the entire field of mathematics for 275 years.

No, but I can prove that the teacher is unqualified to be teaching in school. As evidence, I present her facebook page.

And if you did know the answer, you would keep your mouth shut, write it up in a paper, and get yourself published in a mathematics journal so she wouldn’t steal your credit.

Lol, it would be awesome if the proof to the Goldbach’s Conjecture is solved in the comments section of SandW! Sandra should get 5 points simply for knowing about it.

Also, look up sexy primes. It’s a thing!

No one in today’s strip, and (almost certainly) none of the readers, can prove or disprove it.

But Yuna might.

@ Edda:What makes you think she hasn’t?

well…

First of all you can immediately see that since an odd plus an odd equals an even the only odd numbers you can make are 2 more than a prime, and that after 4 you would never use 2

If you were to try to use a theoretical “base prime”(discounting 2) in a binary fashion, combining sets of two using sets of highest difference the combination method follows the same base pattern as primes themselves, when interpreted as a function, as long as there is no gap in primes larger than that prime multiplied the number prime it is(including one) which is already proven cannot happen, thus the elements in the list of sequential even numbers that can be expressed starting at 2 has a direct ratio to the list of prime numbers, which is proven to go on forever thus the list of sequential even numbers starting at 2 must go on forever! THUS THE ANSWER IS YES!

*drops mic*

@ Pig Lord:But proofs can’t have exceptions (ie 2), that’s why this one is unsolved ATM.

@ Edda:Yuna solving it doesnt count. She’d simply rewrite existance to MAKE it true or false.

There are an infinite number of even numbers and and infinite number of prime numbers in this world that can be combined into an infinite number of combinations to equate to those even numbers thus proving Christian’s claim. Plus I did the math up to 300 also proving my resolve.

@ TvTropesgotmehooked:That can be handled by writing the “2” case as a separate case. For example for something I’ve been working with lately, the standard Gauss-Lucas theorem proof handles the case where for some x f(x)=f'(x)=0 separately from when f'(x)=0 but f(x) !=0.

on the other handle an algorithm is not a proof either, so he’s still not proven it, of course.

Interestingly the related idea of “Is every number is a sum of primes” is trivial. The conjecture has a problem because it restricts it to two primes.

@ Grayson Judd:It’s actually by computer been done up to 2^32.

And apparently the probability is on the order of ^-150.

Cool, how the “it’s so easy, I’ve proved it”-guys are even reading this comic! 😀

As a mathematician one sometimes meets these people at major conferences. They typically stand in front of the entrance and show every mathematician (who isn’t fast enough to escape) their “proofs” of unsolved (or sometimes solved), famous math problems. And of course they are not bothered if the mathematicians point out the flaws in their proofs, because that’s just since they’re ignorant… 😉 Sometimes they are also accompanied by other interesting people, e.g., some who protest against the world domination of mathematics or something. They all would make great psychological case studies, I guess…

But serious, guys, trust me: The Goldbach conjecture is indeed still unsolved. And that’s not because mathematicians are bad at math.

The quiz question, however, can be easily solved. The correct answer is “No, I can’t.”

Byzantinewrote:Where “probability” has to be defined here: at least the usual definition where you take samples doesn’t quite work, since it is always either right or wrong! (Or not provable, but that’s for the experts and a different story.)

@ Nobody:I didn’t actually think i figured it out, and like 20% of that was bs, I just figured i’d be able to learn some cool math that might help with my studies from peoples responses, and I was NOT disappointed!.

Calvin’s answer would probably be a drawing of a spaceship.

Stop it, please! They’re already dead!

@ Mary:But I really need the extra credit.

Eddawrote:I wouldn’t be surprised if Yuna is able to successfully divide by zero.

@ Byzantine:No need to handle 2–the conjecture specifically states all even numbers greater than 2.

Mechwarriorwrote:I wouldn’t be surprised if Yuna counted to infinity. Twice

@ Byzantine:Hobbeswrote:LOL, stop, please! 🙂

Loves these styles of jokes…

Actually, for realism, the math teacher ought to have written this as the bonus question:

“Can every odd number greater than 7 can be expressed as the sum of three odd primes?”

Because it has been proven. Huehuehuehuehue

The proof is optimus. Bugger combines with Everything and is a prime.

Mechwarriorwrote:Dividing by 0 is relatively easy (the answer is always infinity, as I recall – feel free to check it out and correct me) – The difficulty is getting computers to accept that.

And record values like ‘Infinity’ in the first place. Hence NaN.

The fourth exercise in the third edition of Knuth’s “The Art of Computer Programming” is marked as worth 45 points:

“Prove that when n is an integer, n > 2, the equation x^n + y^n = z^n has no solution in positive integers x, y, z.”

In earlier editions, it was 50 points, and might have been worded slightly differently.

Eek, maths.

Mrs G is in a *really* bad mood.

Also, in the fifth panel Sandra correctly notes that the bonus question — “Can you prove or disprove” — is *about* an unsolved problem. It is specifically not *the* unsolved problem.

TvTropesgotmehookedwrote:Proofs can have exceptions, so long as they’re finite in number, and you can prove you’ve identified all of them…

You can disprove by finding a counter example. I looked at a list of primes less than 100 and found you could not create 27 this way. Are you sure the problem isn’t to add or subtract two primes to get the number? That would be harder.

Ok classic read problem more closely. It specifically says even numbers

i’m no expert mathematician (nor a regular one), but i have been wondering something when i was in high school…

since

0÷X=0

X÷X=1

X÷0=∞

would that mean that [ 0÷0 ]’s answer would be all three of those? or does it only count from the divisor to make the answer ∞?

@ Chris:That would technically still be wrong because you’d have to prove that it’s not possible to prove or disprove the theory, which obviously hasn’t been done since it’s still an unsolved theory.

@ Carthienes:Dividing by zero in R is neither easy nor hard, it’s simply undefined.

Extending R to give a meaning to a division by 0 isn’t hard either: you can choose pretty anything you want (for instance I could define a new set Rdumb such that Rdumb contains all elements of R and 1/0=42, it’s completely useless). There are however two pretty common ways of extending R:

– add an “Error” element (called NaN for floating point numbers for instance) and complete each binary operation by saying that it outputs “Error” if it’s otherwise not defined (effectively making it an absorbing element for each binary operation on that set).

– add ∞ and -∞ elements, then define 1/0=∞ and add some other “natural” properties (for instance n+∞=∞, and x*∞=∞ if x>0), this is called the Extended real number line ( https://en.wikipedia.org/wiki/Extended_real_number_line ). That alone doesn’t always define divisions by zero though: 0/0 is still undefined, but it’s an intuitive extension.

There isn’t really any difficulty related to “divisions by zero”: it’s meaningless therefore undefined.

I didn’t know teachers were allowed to use weapons of math destruction.

@ a wandering cynic:0/0 can indeed be anything.

Consider this:

y = (a*x)/x : What is the value of y for x = 0?

You might say the x’s cancel each other out, thus leaving y=a. Technically though, you are still trying to divide 0 by 0 with the result being a. Since a can be any number you like, the result of 0/0 can be anything. For that matter, infinity / infinity, infinity*0 or infinity – infinity also depend on the circumstances. It’s a delicate matter.

Some more 0/0 problems:

sin(x) / x for x =0

(1 – cos(x)) / x for x = 0

square root of x / sin(x) for x = 0

Some infinity / infinity problems:

x^n / e^x for x -> infinity

ln(x) / (1/x) for x = 0

Carthieneswrote:IEEE floating point encodes and handles infinity just fine. 1/+0 = +inf. -1/+0 = -inf. log(0) = -inf. Try it out in your favourite programming language or calculator. Some languages/calculators specifically check for division by zero and produce an error, but it’s a deliberate choice intended to make debugging easier made by the designers of those languages, not a limitation of the underlying floating-point representation. For example, Python does this, but you can still get infinity by explicitly writing

`float('inf')`

or by performing the division by zero in NumPy’s`numpy.float_`

type instead of the built-in`float`

.NaN occurs for indeterminate forms such as 0/0 and inf/inf, as well as when there is literally no valid extended-real value, e. g. in

`log(-1)`

and`sqrt(-1)`

.I was careful to pick Yuna as the only possible “solver”. Anyone who can reverse entropy…

S&W also features assorted gods and demons as characters. You’d think they’d know. But, except for Seeoahtlahmakaskay, they all seem pretty incompetent. And the raccoon-goddess probably doesn’t care.

Eww. I know I should’ve looked at the HTML tag examples below the comment submission form before submitting mine, but Novil, <code> is an inline element, and it would be awesome if your stylesheet didn’t betray this expectation. For code blocks, <pre><code> is used.

In the transcript when Sandra is saying … Can you prove or disprove Christan’s claim, it is in German.

@ Asrial:*slaps* get out of here

No group, or generalization thereof, can contain division by zero. As such, any “extended real numbers” is not a group, field, module, or anything from that family.

The reason is simple: If you define t=1/0, then you must necessarily also define t*0=1, since that’s part of the definition of division. Then, for an arbitrary number a, you can receive a=a*1=a*(t*0)=(a*t)*0=(a*t)*(b-b)=(a*t*b)-(a*t*b)=0. The latter is part of the definition of 0 in a group.

So, for this to work, you either need multiplication to be partially undefined, and/or for it to not adhere to basic rules like associativity, and/or division to be meaningless in relation to multiplication.

Contrast with i=sqrt(-1), in which i*i=-1, and no contradictions occur (multiplication and addition are still fully defined).

Incidentally, when it comes to limits, division by 0 IS defined, but there is no single 0. That is, 1/0 can be infinity, but it can also be negative infinity. That’s because you can have positive 0, and negative 0. You can also have sign-changing 0, causing 1/0 to be undefined. People who think you can simply “add infinity” are missing this important point.

As for Goldbach, if Knuth’s intuition that it is undecidable in PA is valid, then it would simultaneously be true, and yet Sandra’s correct answer would be “no”, since a proof that doesn’t rely on PA would be invalid in school.

Alternately, if all possible proof systems are allowed, than the answer would be “yes”. Trivially, if it is not true, then the counter-example can be shown. If it is true, it can be added to PA, resulting in a system with no contradiction, and in which the proof of Goldbach is trivial (by axiom). Ergo, it necessarily can either be proven, or disproven.

Sandra: … Kannst du Christians Behauptung beweisen oder widerlegen? What the hell is this?

@ 123456:TvTropesgotmehookedwrote:Well in essence the real question is devising and proving an analytical formula to devise nth primal number.

If I had that it would be easy to prove or disprove by induction.

@ Paris Kiran:Such formula may not exist, making analytical proof impossible.

@ Nobody:Acutally it can be more easily solved, since what the teacher wrote is not the Goldbach conejecture. Since the teacher used “number” instead of integer, one could grab an rational number such as 2.5, since 2 is the only prime lower and all other number that may be added are lower that one, and as such there are no “prime” (since the primes are only correctly defined in the integers, I’m using the quote marks) that fits the bill. Q.E.D.

Now if she asked for all integers, that would be Goldbach’s conjecture and as such all that is being said here irrelevant, but not defininfg the problem can make it far easier.

The English transcript has a German line:

Sandra: … Kannst du Christians Behauptung beweisen oder widerlegen?

@ kurokotetsu:

Prime elements are meaningful not only in the positive integer numbers but in any integral domain: https://en.wikipedia.org/wiki/Integral_domain#Divisibility.2C_prime_elements.2C_and_irreducible_elements

For example 5 is prime in Z but not in Z(sqrt{5}) (the set of numbers a + b*sqrt(5) such that a and b are integers where a square root is obviously 1*sqrt(5)).