EPTThe Math Thread - Page 27
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Deleted User 3420
24492 Posts
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Melliflue
United Kingdom1389 Posts
On April 17 2019 06:13 travis wrote: oh yeah im dumb its just k-->2^k, right? Yep. 2^j x 2^k = 2 ^(j+k) so it is a group homomorphism. Btw, if a map f between groups satisfies f(gh)=f(g)f(h) for all g,h then it must map the identity to the identity since f(1)f(g)=f(1g)=f(g). | ||
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Deleted User 3420
24492 Posts
I have another question, to compute 101^(4,800,000,023) (mod 35) by hand I honestly can't figure out how to do this one. I had a feeling, so I checked 101^10 mod 35 and saw that it is 1. So I can do 101^23 mod 35 to get my answer. I can just do that by using powers of two. But... I only know that 101^10 mod 35 is 1 because I checked with wolfram alpha (I had a suspicion). By hand, it's not like I would just be checking every power.. Icouldn't. So how am I supposed to do this one? | ||
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mahrgell
Germany3943 Posts
On April 17 2019 21:41 travis wrote: ok cool I have another question, to compute 101^(4,800,000,023) (mod 35) by hand I honestly can't figure out how to do this one. I had a feeling, so I checked 101^10 mod 35 and saw that it is 1. So I can do 101^23 mod 35 to get my answer. I can just do that by using powers of two. But... I only know that 101^10 mod 35 is 1 because I checked with wolfram alpha (I had a suspicion). By hand, it's not like I would just be checking every power.. Icouldn't. So how am I supposed to do this one? ab mod n = (a mod n) (b mod n) mod n a^b mod n = ((a mod n) ^ b) mod n a^(b+c) = (a^b)(a^c) a^(bc) = (a^b)^c Those rules are enough to do it in only few steps. | ||
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Simberto
Germany11913 Posts
This is 31 or -4 Then you just calculate the first few powers of 31 that a bit to see when it ends up at 1. You will at most ever need the modulo number in steps due to stuff that you should have proven in group theory at some point. You can then simply remove any multiples of this number from the exponent without problems, as they are 1s. Solution: + Show Spoiler + ^1 : -4 ^2 : 16 ^3 : -64 = -29 = 6 ^4 : -24 = 11 ^5 : -44 = -9 ^6 : 36=1 (I might have calculated something incorrectly here, because that means that 101^10 is not 1 (mod 35) You can then simply remove any multiples of this number from the exponent without problems. So in your case, you can remove 4800000018 from the exponent, because that is dividable by 6 Meaning your result is 31^5 mod 35, which we already know is -9 = 26 due to the calculations we did above. | ||
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enigmaticcam
United States280 Posts
// Calculate x^y % z | ||
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Deleted User 3420
24492 Posts
On April 17 2019 23:47 Simberto wrote: Basic idea would be to first calculate 101 mod 35, because there is no reason to ever have a number above 34 if you are calculating mod 35 anyways. This is 31 or -4 Then you just calculate the first few powers of 31 that a bit to see when it ends up at 1. You will at most ever need the modulo number in steps due to stuff that you should have proven in group theory at some point. You can then simply remove any multiples of this number from the exponent without problems, as they are 1s. Solution: + Show Spoiler + ^1 : -4 ^2 : 16 ^3 : -64 = -29 = 6 ^4 : -24 = 11 ^5 : -44 = -9 ^6 : 36=1 (I might have calculated something incorrectly here, because that means that 101^10 is not 1 (mod 35) You can then simply remove any multiples of this number from the exponent without problems. So in your case, you can remove 4800000018 from the exponent, because that is dividable by 6 Meaning your result is 31^5 mod 35, which we already know is -9 = 26 due to the calculations we did above. err yeah sorry, it was 101^10 mod 35 was 11 not 1 your solution is probably exactly right, ill verify it and I think I could see that to reach the identity we would need to raise to a power of a number that divides 24 right? using euler's theorem, we can take 35 = 7*5 = order of (7-1)*(5-1) = 24 I think this is correct application? EDIT: after going through your solution I see what you did, you used the -4 all the way through... that's really cool lol | ||
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Simberto
Germany11913 Posts
So you can also just remove any multiples of 24 from the exponent without even doing any work whatsoever. That still leaves the 23 as an exponent, which is equal to ^-1. So you just need to find the inverse of 31, if you want to go about it that way. But looking for inverses in modulo groups was pretty annoying and basically involved just testing all group members until you find one that works, if i recall correctly. | ||
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Mafe
Germany5966 Posts
On April 18 2019 02:19 Simberto wrote: + Show Spoiler + Yes, that sounds right. You need to verify that 35 and 101 are coprime for this to work, which they are. So you can also just remove any multiples of 24 from the exponent without even doing any work whatsoever. That still leaves the 23 as an exponent, which is equal to ^-1. So you just need to find the inverse of 31, if you want to go about it that way. You can use the extended euclidean algorithm to do that. | ||
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Deleted User 3420
24492 Posts
It says, Suppose G is an abelian group and α : G --> G is defined by α(G) = G^2 a.) show that α is a homomorphism. b.) identify ker α and α(G) in the case where: 1.)G = U(11) and 2.)G = U(15) for part a I put α(a*b) = (ab)^2 = a^2b^2 = α(a)α(b) correct? and then for part B i don't even understand how to answer. Am I listing out the case for each member of G? | ||
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Simberto
Germany11913 Posts
In b, you need to understand what "ker(alpha)" and alpha(G) means. Those are not statements that talk about single elements of the group. the ker is everything that maps onto 1 via this homomorphism, and the image alpha(G) is everything that you can reach by starting with an element of G and using this homomorphism. And yes, basically the easiest way to get this is to simply calculate alpha of each member of G, there are not that many in both groups anyways and alpha is not that hard to calculate. Figure out which go onto 1, and which of the members of G are actually something that can be reached via this homomorphism. Those are your two answers. Edit Example: 1*1=1 ===> 1€ker(alpha) and 1€alpha(G) 2*2=4 ===> 2 not€ker(alpha) because it doesn't map onto 1 and 4€alpha(G) because 2 maps onto 4 (In both groups) | ||
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Deleted User 3420
24492 Posts
question for everyone: anyone around these parts have a good grasp of circumscription? (like in logic and reasoning, for an AI class). I can't make heads or tails of it, we were given little resources with which to learn it and I find the notation in the homework to be incredibly confusing | ||
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brian
United States9642 Posts
what are the odds that in two million coin flips, i have a result of 54% or greater in favor of heads? (and specifically not tails, though that would be a follow up question. what if i wanted 54% or greater in either direction, is it just twice the probability?) | ||
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Acrofales
Spain18348 Posts
On April 25 2019 01:44 brian wrote: i have a dumb question that i couldn’t google. what are the odds that in two million coin flips, i have a result of 54% or greater in favor of heads? (and specifically not tails, though that would be a follow up question. what if i wanted 54% or greater in either direction, is it just twice the probability?) Knock yourself out: https://stattrek.com/online-calculator/binomial.aspx | ||
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brian
United States9642 Posts
On April 25 2019 01:54 Acrofales wrote: Knock yourself out: https://stattrek.com/online-calculator/binomial.aspx nice! but, it only goes to 100,000 trials  , and i don’t understand how it would tell me the odds of getting 54% or better i guess. it’s been a decade since i’ve been into math. oh i can see, i put in 54,000 success and take the probability of it being greater. sweet. halfway there. | ||
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enigmaticcam
United States280 Posts
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Melliflue
United Kingdom1389 Posts
On April 25 2019 01:44 brian wrote: i have a dumb question that i couldn’t google. what are the odds that in two million coin flips, i have a result of 54% or greater in favor of heads? (and specifically not tails, though that would be a follow up question. what if i wanted 54% or greater in either direction, is it just twice the probability?) For that many coin flips I think you can safely use a normal distribution because of the central limit theorem. Make a head 1 and a tail 0 so that the final number is how many heads you got. But I think the answer will be so close to zero as to be negligible. Btw, you are correct that it is twice the probability because in general P(A or B) = P(A) + P(B) - P(A and B) and if A is "54% or more heads" and B is "54% or more tails" then P(A and B)=0 because you cannot have both 54% or more of heads and tails. By symmetry P(A)=P(B). | ||
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Acrofales
Spain18348 Posts
On April 25 2019 02:12 enigmaticcam wrote: Question on something I've been struggling with. If I have a line with a slope of x, and this line reflects off a mirror with a slope of y, is it possible to calculate the slope of the reflected line using basic algebra without trig? Yes. You can use an affine transformation matrix. Specifically, you can use the Householder transformation: https://en.wikipedia.org/wiki/Householder_transformation E: just saw the wikipedia page for that is needlessly broad. Just use this: https://en.wikipedia.org/wiki/Transformation_matrix#Reflection | ||
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Ciaus_Dronu
South Africa1848 Posts
On April 25 2019 02:12 enigmaticcam wrote: Question on something I've been struggling with. If I have a line with a slope of x, and this line reflects off a mirror with a slope of y, is it possible to calculate the slope of the reflected line using basic algebra without trig? It requires a bit of 2D geometry, but yes. This stack exchange post, the second answer in particular, explains how: Stackexchangeislove Remember that the 'normal vector' to the 'mirror line' has slope -(1/y). | ||
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Simberto
Germany11913 Posts
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