At the Airspeed Velocity of an Unladen Swallow

floateron:

mynerdproblems:

hellicaptor:

mycroft-queenofcake:

iamjayse:

thenerdfighterkid:

slydig:

tsarbucks:

slydig:

dont be mean 

be median or mode

damn math fandom bloggers

shut up we have a good range of jokes

this is our domain

guys we’re forgetting the point of this post and going off on a tangent

well i for one think math puns are pretty radical

math puns are the first sin of madness 

cos then the imaginary starts seeming real

slydig:

tsarbucks:

slydig:

dont be mean 

be median or mode

damn math fandom bloggers

thewritingfortress:

aikomaru:

theoddginger:

aikomaru:

if school taught everyone that rape is bad and racism is bad instead of teaching how to find the angle of a kite im pretty sure the world would be a safer place

except all the kites would have wrong angles and spontaneously fall and kill people

i think youre missing the point buddy

I’d like to hear you say that when your impaled by a falling kite that’s shaped incorrectly. 

fictionboyfetish:

grouprojects:

titmuffins:

grouprojects:

being gay is a sin?? um actually

bi = gay x straight

sin = straight / bi

the straights cancel

sin = 1/gay

times that by gay

gay(sin) = 0

gay = -sin

move the negative over

-gay = sin

so not being gay is a sin oops

idk what you just said but thanks nerd

your math is atrocious hold on
sin is not an expression you use like that it’s a function you need an input
so we’re talking about the sin of being gay

I’ll agree that bi = straight x gay, but what you need is

sin(gay) = opposite / hypotenuse

sin(gay) = straight / bi

sin(gay) = straight / (straight x gay)

sin(gay) = 1/gay

now we take the arcsin to undo the sin

arcsin(sin(gay)) = arcsin(1/gay)

gay = arcsin(1/gay)

it is jesus who absolved our sins so the arcsin function is also the jesus function

gay = jesus(1/gay)

this is the unit circle:

If we reconsider the x-axis as the kinsey scale then we can say that the point (-1, 0) is fully straight and (1, 0) is fully gay, so

gay = jesus(1/gay) = jesus(1/1) = jesus(1)

there is one god, the father the almighty, creator of heaven and earth

gay = jesus(god)

in english: “being gay is equal to jesus of god,” or phrased more clearly

Being gay is viewed as being equal to any other person in the eyes of Jesus, the son of God.

QED.

ಠ_ಠ

tupacabra:

forgive me father trigonometry, for i have sined

studentrndofficial:

Since elementary school, we learned basic mathematics skills as little children. As we grew older, our math improved as we learned new concepts. Yet have people ever wondered why Americans lag behind Eastern Asian countries, such as China, in math? The answer might not easily be what you think:

The answer lies not only in the practice that Asian students receive but also, surprisingly, in the language we speak. Examine the following numbers: 8,2,4,6,7,5,1. Now look away for twenty seconds, and try to memorize the order of the numbers presented. Research has shown that you have a 50% chance of accurately memorizing that sequence perfectly, if you speak English.

Read More

mathmajorproblems:

aoiasahina:

(p + l)(a + n) = pl + pn + la + ln

i foiled your plan

DED

thatmathblog:

sundalius:

thatmathblog:

Every gambler wishes there was a way to be guaranteed infinite money. Even better so if it goes against expectations — using losing strategies. 

Enter Parrondos’ paradox, which says just that. Using the stochatisc process classes’ commonplace example of biased coin games, one can devise a strategy for alternating between 2 games, A and B, which guarantee wealth — despite a guaranteed bankruptcy playing A or B separately. 

For game A, let there be a biased coin, P, where the probability of winning on P is p=0.5-x. Winning gains you a dollar, and losing loses you a dollar. After each game of A, examine your current bankroll — if it is a multiple of 3, play biased coin Q; else, play biased coin R. For coin Q, the probability of winning is q=0.75-x; for coin R, the probability of winning is r=0.1-x. The winnings are, once again, a dollar and losing subtracts a dollar.

In other words, you have a 50% chance of winning game A and either a 10% or 75% of winning game B. 

When played alternately, the end result is a seeming paradox — that the limit of the money quickly ascends into infinity! (note that not ALL alterations of the game are winning — playing  ABABABAB…. is a losing game, while ABBABBABBABB….is winning). 

Why? Let’s look a simpler example (since I’m NOT about to type out transition matrices 3897234 times):

1. In A, you lose 100% of the time and each time, you lose $3. 
2. In B, you win $8 if your bankroll is a multiple of 5 (0 counts as a multiple of 5), and lose $6 if it is not. 

Let game A just be throwing your money into a blender. Just chuck $3 right in. Game B, if you have any number of $5 bills, you get $8. Because why not. Otherwise, they take $6 and poop on it, THEN throw it in the blender. Because fuck you, that’s why.

Let c(t) be the bankroll at game instance t, and consider c(0)=m. We will play the games alternatively ABABAB….

Suppose m will never reach a multiple of 5 for all t. This means that m and 5 are coprime. Suppose you play A then B — this gives you m-3-6=m-9, which must also be coprime with 5 (since we are assuming that m is never divisible by 5 at any time). So, suppose there are natural numbers a,b,c,d, for 0< b,d <5, such that m=5a+b and m-9=5c+d by the division algorithm. Generalizing this for any time t:

m-9(t)=5(c-2t)+d+2t
m-9t=5c-10t+d-2t
m-9t=5c+d-12t. 

Now, since 0<d<5, we can explore all the possibilities as thus: 

d=1; m-9(t)=5c+1-12t —> m=5c+1-12t+9t=5c+1-3t. Suppose this were coprime with 5 for all t. Then 1-3(t)=\{1,-2,-5,…\} and we can stop — t=2 says m=5c+1-3(2)=5c+1-6=5c-5=5(c-1), which is a multiple of 5. Contradiction.

d=2; m=5c+2-3t —> 2-3t=\{2, -1, -4, -7, -10,…\}, and again, we stop; t=4 —> 2-3(4)=-10, so m=5c-10, which is divisible by 5. Contradiction.

I’ll leave the rest up to the reader, since they’re all the same proof.

As this shows, no matter what your starting capital is, (well, you need $8 or more, since only $3 gets you the infinite sequence of 5 multiples needed), you will always gain infinite money!

Too bad the casinos know about this stuff.

Isn’t there an error where it says m-3-6=m-9 must be co-prime to five, as it uses this equation to balance out others, where they use it beyond m=m, since both sides are m-9=m-9 when simplified.

Still a good read.

The assumption is that there exists a number that will always be coprime to 5 for all t. Since it will always be coprime to 5, for all t, we see that it must never gain 8 — only lose 6 (during game B). As such, during game A, we subtract 3, then game B, subtract 6. This is only used to set up the general case. From there the general case is analyzed more in the context of the division algorithm, which does mean that they need to be equal, but we need the more specific….”components” of the equality. For example, 13=2(5)+1=2(6)-1…. etc. Without them, we cannot explore the possibilities for d, which ultimately lead us to the contradiction — there cannot be a number that’s never coprime to 5 through this game system (but we don’t deal with negative monies). Hence, it’s necessary to set it up as we did.

If this helped, good; if not, shoot me a message with what you don’t understand and I’ll try to explain!

What’s amazing about Physics and Mathematics is that it gives you a satisfying feeling when after the nerve cracking equations and long, blood-sweating solutions, you finally get the right answer.

It’s like courting a girl and after long months or years of hard work and patience you finally get to hear that sweet YES.