/sci/ - Science & Math

[Anonymous No.2369036]

/sci/ You're good at math soo you're probably good at discrete math too.

Show that all prime numbers can be written as 6n+1 or 6n-1.

pic unrelated

[Anonymous No.2369040]

busy looking at tits...go away.

[Anonymous No.2369041]

>all prime numbers
2, 3?

[Anonymous No.2369043]

How about all prime numbers greater than 5 as 6n+1 or 6n+5?

[Anonymous No.2369048]

Yeah.
You're suppose to reward AFTER the trick is done, not before.

[Anonymous No.2369054]

>>2369043
>6n+5?

[Anonymous No.2369056]

>>2369048
You're forgetting the very first rule of bartering, arguing etc.: You always start low, hoping they'll take that initial offer as a sign that you have a lot more money, or in this case a lot better pictures.
When he gets what he wants I bet he'll post more (hopefully), how much more varies from person to person.

[EK !D96TK2PSsI No.2369061]

/sci/ is meant to be SFW board OP, but w/e.

also, i don't know how to explain it mathematically, but all prime numbers bigger than 2 or 3 must be in the form 6n+1 or 6n-1 because all other numbers must be divisible by 2 or 3, and hence are not prime.

6n (divisible by 2, 3 and 6)
6n+1 (prime)
6n+2 (divisble by 2)
6n + 3 (divisible by 3)
6n + 4 (divisible by 2)
6n + 5 (prime)

[EK !D96TK2PSsI No.2369064]

>>2369048
>implying tits are any sort of reward, when we are on the internet and merely googling 'tits; with safesearch turned off yields this 'reward'

>>2369054
6n + 5 can be seen as equivalent to 6n - 1. its just that n is 1 number higher in the second case.

[Anonymous No.2369070]

>>2369061
>6n + 5 (prime)

n = 5
6x5 + 5 = 35

[EK !D96TK2PSsI No.2369078]

>>2369070
EDIT: 6n, 6n + 2, 6n +3 or 6n + 4 CANNOT be prime because they MUST be divisible by 2 or 3 regardless of what n is.

6n +1, or 6n + 5 MIGHT be prime, but not always.

[Anonymous No.2369079]

>>2369070

Show that all prime numbers can be written as 6n+1 or 6n-1.

=!

all 6n+1 and 6n-1 are prime

[Anonymous No.2369081]

>>2369070
You don't understand the question, the correct statement is for all prime p>3, <span class="math">p \mod 6 \in \{1,5\}[/spoiler], which is fairly straightforward.

[Anonymous No.2369085]

>>2369056
When you make a low ball deal you know it's bad. You make it to see if they'll take it (You're praying they're ignorant of the value or desperate). There's a lot of people unintelligent about what they're selling (not stores usually, people. Although the person working at the store could be mentally fucking gone for all you know. He's working at a store, you don't have to worry about running into a fucking wordsmith genius. Unless you're dealing with chemicals/pharmacists etc
Every situation is different, though. So you will almost always have to adjust your approach, and tactics. If you make a person who knows the value a bad offer, they'll most likely raise it a little bit higher than if you had given them a decent, fair offer (but still low to you. You want the trades / sales to benefit you, not them.)

[EK !D96TK2PSsI No.2369087]

all squares are rectangles, but not all rectangles are squares.

all primes can be written in the form 6n+1 or 6n-1, but not everything that can be written in the form 6n+1 or 6n-1 is prime.

[Anonymous No.2369088]

the statement is equivalent to saying every prime > 5 is congruent 1 or -1 mod 6.
Hm.

If p is a prime, it can't be 0 mod 6.
it also can't be 2 mod 6, since then it would be even.
for the same reason it can't be 4 mod 6.
it can't be 3 mod 6 since then it'd be divisible by 3.

Thus it must be either 1 or 5 = -1 mod 6.
6+1 = 7 and 2*6 - 1 = 11 are primes, thus there are examples for both.

q.e. fucking d.

[Anonymous No.2369095]

>>2369070
ITT: Sean Connery Primes

>77

[i- !MccPZRjodM No.2369097]

>>2369054
wow you are retarded

[Anonymous No.2369101]

>>2369088
i think EK pretty much already said the same thing

[Anonymous No.2369102]

>>2369088
OP here, VERY, FUCKING, NICE. It's perfect.

[Anonymous No.2369105]

This thread reasserts my long standing view that Mathematics for Mathematics' sake is utterly pointless bullshit. See below:

>all primes can be written in the form 6n+1 or 6n-1, but not everything that can be written in the form 6n+1 or 6n-1 is prime.

What the flying fuck is the purpose of proving this lousy statement

[Anonymous No.2369107]

>>2369101
I don't usually read EK's bullshit.
But thanks for letting me know.

[Anonymous No.2369108]

>>2369107
waaaaaaaaaahhhhhhhhhhhhhhhhh

[Anonymous No.2369130]

>>2369105
>implying primes arent useful

[Anonymous No.2369131]

Here's a good one:

the product of any n consecutive integers > 0 is divisible by n!.

[Anonymous No.2369135]

123456789

That is all.

[Anonymous No.2369157]

>>2369130
That's not what I was implying at all, and you know it. I am attacking the idea of proving roundabout statements, such as the one I quoted previously.
Since when does a inconsistent set of equations, with no conditions given, benefit anyone?

[Anonymous No.2369164]

>>2369157
what in the name of dick are you talking about?

modular arithmetic is used to the point of no return in cryptography. So please, go away.

[Anonymous No.2369171]

>>2369157

Knowing this statement, you can immediatly make a program that spits out primes, and even with a bruteforce algoritm (n=1,2,3..) and a prime check it will still be faster than checking every integer.

Might be useful in some obscure setting, or it might be a good drill in learning to prove simple statements from thinks you know. How can you group logical statements in two discrete groups: (useful) and (not useful) ?

[Anonymous No.2369176]

I'll try to hijack this and post another question about induction. Use induction to prove that (refer to pic) is correct. where n is a positive whole number.

[Anonymous No.2369189]

>>2369176
:<

[Anonymous No.2369193]

>>2369176
insert n+1

[Anonymous No.2369209]

>>2369193
Yeah but what about the sum on the left side, no idea how to handle it? :S

[Anonymous No.2369233]

>>2369209

what happens to the fractions for large n?
maybe you can replace it with simpler terms

[Anonymous No.2369240]

>>2369157
You're either trolling or an idiot, and I've replied to far more transparent trolls today already. What you're saying is essentially that statements of the form "all x are y, but not all y are x" are useless, which is blatantly false - they're called implications and form the foundation of any logical argument. Knowing that every prime has a certain property is useful even if not every integer with that property is a prime. For instance, Fermat's little theorem endows primes with a certain property which is also held by plenty of other integers. However, knowing this enables us to write a far more efficient prime-checking algorithm than most other equally elementary approaches.

[Anonymous No.2369248]

>>2369176
You know how the left-hand side changes when you increase n by one. Just show that the right-hand side increases more, by calculating how much it does.

[Anonymous No.2369279]

>>2369176

also, not strict unequality for n = 1

[Anonymous No.2369284]

By the division algorithm, all integers can be uniquely written in the form <span class="math">6q+r[/spoiler] where <span class="math">q,r[/spoiler] are integers and <span class="math">0\leq r<6[/spoiler]. <span class="math">6q+0[/spoiler], <span class="math">6q+2[/spoiler], <span class="math">6q+3[/spoiler], and <span class="math">6q+4[/spoiler] can never be prime because they are divisible by <span class="math">2[/spoiler], <span class="math">2[/spoiler], <span class="math">3[/spoiler], and <span class="math">2[/spoiler], respectively. Thus all primes must be of the form <span class="math">6q+1[/spoiler] or <span class="math">6q+5[/spoiler]. Now observe that <span class="math">6q+5=6(q+1)-1[/spoiler], so the form <span class="math">6q+5[/spoiler] is equivalent to the form <span class="math">6q-1[/spoiler].

[Anonymous No.2369287]

>>2369284
Excluding <span class="math">2[/spoiler] and <span class="math">3[/spoiler], of course.

[Anonymous No.2369288]

>>2369240
That's a whole lot of words for a mathematician. You should stick to arguing by numbers, my man. You know where you are with numbers

[Anonymous No.2369297]

>>2369248
if you added +1 to just k/(k+1) you'd have n+1/n+2 on the left side, on the right side you'd have (n+1)^2/(n+2). I'm just struggling trying to understand how you can write how the sum and the +1 is still smaller than (n+1)^2/(n+2) when all you know is that n^2/n+1 is bigger than the sum?

[Anonymous No.2369302]

>>2369176
there is probably some cute trick to it, but you could say that the second term minus the sum is bigger than zero.

express this a function and show that it is monotonly increasing(or what it is called in english)

[Anonymous No.2369303]

>>2369070
All prime numbers can be expressed in that form, not that all possible values of n make a prime number.

[Anonymous No.2369393]

>>2369288
sadly, you have already been proven an idiot long ago

see:
>>2369164

any other post coming from your ass will be equivalent to a 5 year old with downs arguing with himself

[Anonymous No.2369506]

>>2369176
Thanks for the fun problem. Certainly you meant <span class="math">n-1[/spoiler] on top of that sum. Anyway, hopefully I have no typographical errors.

Claim: <span class="math">\sum_{k=1}^{n-1}\frac{k}{k+1}<\frac{n^2}{n+1}[/spoiler] for positive <span class="math">n[/spoiler].

Proof: We proceed by induction. Clearly the claim holds for <span class="math">n=1[/spoiler]. Suppose it holds for some fixed <span class="math">n[/spoiler]. Then

<span class="math">\sum_{k=1}^{(n+1)-1}\frac{k}{k+1}[/spoiler]
<span class="math">=(\sum_{k=1}^{n-1}\frac{k}{k+1})+\frac{n}{n+1}[/spoiler]
<span class="math"><\frac{n^2}{n+1}+\frac{n}{n+1}[/spoiler] (by the induction assumption)
<span class="math">=n[/spoiler] (an impressive simplification)
<span class="math">=n\frac{n+2}{n+2}[/spoiler] (multiplying by one, here comes the magic...)
<span class="math">=\frac{n^2+2n}{n+2}[/spoiler]
<span class="math"><\frac{n^2+2n+1}{n+2}[/spoiler]
<span class="math">=\frac{(n+1)^2}{(n+1)+1}[/spoiler],

i.e. the claim holds for <span class="math">n+1[/spoiler]. QED.

[Anonymous No.2369562]

>>2369506
Ah, my apologies, I misread at first; <span class="math">n[/spoiler] on top of the sum is fine, I'll fix this up.

[Anonymous No.2369564]

>>2369176
why the hell would I use induction ?

<div class="math"> \sum_{k=0}^n \frac{k}{k+1} = n - \sum_{k=0}^n \frac{1}{k+1} < n - \frac{n}{n+1} = \frac{n^2}{n+1} </div>

[Anonymous No.2369570]

>>2369564
from k=1 ofc

[Anonymous No.2369712]

>>2369562
Here we go.

Prove <span class="math">\sum_{k=1}^n\frac{k}{k+1}<\frac{n^2}{n+1}[/spoiler] for positive <span class="math">n[/spoiler] by induction.

Clearly the claim holds for <span class="math">n=1[/spoiler]. Suppose it holds for some fixed <span class="math">n[/spoiler]. Then

<span class="math">\sum_{k=1}^{n+1}\frac{k}{k+1}[/spoiler]
<span class="math">=(\sum_{k=1}^n\frac{k}{k+1})+\frac{n+1}{n+2}[/spoiler]
<span class="math"><\frac{n^2}{n+1}+\frac{n+1}{n+2}[/spoiler] (by the induction assumption)
<span class="math">=\frac{n^3+3n^2+2n+1}{(n+1)(n+2)}[/spoiler] (combining terms with a common denominator)
<span class="math"><\frac{n^3+3n^2+3n+1}{(n+1)(n+2)}[/spoiler] (here's the magic: changing the <span class="math">2n[/spoiler] to <span class="math">3n[/spoiler]...)
<span class="math">=\frac{(n+1)^3}{(n+1)(n+2)}[/spoiler] (...so that we actually have <span class="math">(n+1)^3[/spoiler] on top)
<span class="math">=\frac{(n+1)^2}{n+2}[/spoiler],

i.e. the claim holds for <span class="math">n+1[/spoiler].