B
Just a preference, sweaty.What do you base this on?
Credit card billing cycles are jewed.
C
Just unpin this garbage lmao
C
A classic elementary one from Euler, taken from the Big Book of T:
(Paraphrasing since I don't quite remember the problem) (NT) Prove that a pos integer can be written as a sum of two perfect squares iff it can be written as a sum of squares of two rational numbers
(Paraphrasing since I don't quite remember the problem) (NT) Prove that a pos integer can be written as a sum of two perfect squares iff it can be written as a sum of squares of two rational numbers
While this is not too difficult if you're willing to use the sum of two squares theorem, I couldn't immediately find a way to prove it without going that route. It it possible, however, as this very elegant proof shows.
it's been a while since i posted here.
let s(n) be the sum of positive divisors of n.
prove that s(n) < n + nlogn for all integers n >= 2.
[hint] integrate 1/t
let s(n) be the sum of positive divisors of n.
prove that s(n) < n + nlogn for all integers n >= 2.
[hint] integrate 1/t
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Let S(n) be the sum of proper divisors of n. We equivalently show that S(n) < n*ln(n) for n > 1.
S(n) = sum n ≠ k|n of k = n * sum n ≠ k|n of k/n = n * sum 1 ≠ k|n of 1/k ≤ n * sum 1 < k ≤ n of 1/k = n * sum 1 < k ≤ n of (1/n)/(k/n) < n * int 1/n < t < 1 (here we use that n > 1) of 1/t = n * ( ln(1) - ln(1/n) ) = n*ln(n)
dang you're quick good job. i wish we had a LaTex embedd somehow so it woulda been a lot easier to read tho
Thanks. Yeah, I know it reads like ass, but I was too lazy to externally LaTeX it, sorry.
Hi I'm mentally retarded and barely able to grasp high school math, when I approach things like linear algebra and imaginary numbers how do I mentally approach it? Is linear algebra just a description of something's position on a 2d plane so I can visualise it and are imaginary numbers making up theoretical numbers for seemingly ridiculous equations so they work, or am I grasping it wrong? wtf is mathematics even?
The beauty of the more fundamental mathematical concepts (such as linear algebra and complex numbers) is that there are plethorae of ways to construe them. While imaginary numbers were initially indeed conceived to solve cubics (casus irreducibilis) there are -- dare I say -- better ways of thinking about complex numbers. One POV I really like is the following one.
In essence, complex numbers are just 2d vectors with a notion of multiplication (which, unlike the inner product, yields another 2d vector). This notion of multiplication is nice, because it geometrically encapsulates rotations. That said, your description of complex numbers is no less wrong (except that I would argue that the equations they solve aren't "ridiculuous" but details) than the one I just offered.
Your description of linear algebra feels a bit barebones to me, but that's probably because I know a lot more linear algebra than you do. As for delineating mathematics, mathematics is one of those disiplines that's hard to fully capture in words. Philosphy would be another such discipline.
TL;DR these questions are frankly hard to answer.
Heres a fun problem that even someone who doesn't know too much could solve
Take a number such as 128916 (though it works for any whole number)
Add up the digits 1+2+8+9+1+6=27
If the total is less then 9 it is not a multiple of 9, if it is 9 then you know its a multiple of 9, if its more you add up the digits again
In this case 27>9 so we add the digits again (2+7)=9 which proves it is a multiple of 9
The question is how do you prove why this process works?
Hint 1 You can express a number such as: 123 = 100+20+3
Hint 2
1/9=0 remainder 1
10/9=1 remainder 1
100/9=10 remainder 1
ect....
Hint 3
If the total remainder is equal to 9 or any multiple of 9 then you have proven it is a multiple of 9, since it would have no remainder
Answer
Simple Case:
Answer take a number 1323
We can express it as 1000+300+20+3
1000/9= something remainder 1
300/9= something remainder 3
20/9= something remainder 2
3/9= something remainder 3
1+3+2+3=9 is the remainder, since its remainder is 9 that essentially means it has 0 remainder since we are dividing by 9
Extended Case:
Take a number 128196
We can express it as 100000+20000+8000+100+90+6
100000/9= something remainder 1
20000/9= something remainder 2
8000/9= something remainder 8
100/9= something remainder 1
90/9= something remainder 9 (I know it should be 0 but you can technically think of it as remainder 9 since they are equivalent)
6/9= something remainder 6
1+2+8+1+9+6=27 is the remainder is more then 9 we must do it again
Take 27
We can express it as 20+7
20/9= something remainder 2
7/9= something remainder 7
2+7=9 is the remainder which proves that 27 is a multiple of 9
Since we got a remainder of 27 for 128196 and we know 27 is a multiple of 9 that means that 128196 is a multiple of 9, since its remainder is a multiple of 9 (making it 0)
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It works because you have set this condition of "if it's not 9 - recount it" so sooner or later it will become 9.
Wish I were better at mathematics.
N digit number can be written in base 10 notation as
A * 10^(n-1) + B * 10^(n-2) + .. + X * 10^9
We know that 10 === 1(mod 9), this also works for 10^k and can be written as 10^k === 1(mod 9), use mathematical induction to prove it, will cut it short and skip
This rule can be used to reduce our equation to
A * 1 + B * 1 + .. + X * 1
Which means that the sum of original number is congruent to the sum of its digits mod 9, if the number is large you can just reapply the rule as the property holds in each step
In essence this, altho your notation is somewhat sloppy (e.g., using A thru X instead of indexed variables) and you haven't argued why this procedure actually decreases your number at every step (so long as it has at least two digits anyways) altho this is pretty obvious.
For those interested, this process is called taking the digital root btw.
yea fair enough, defs could've been expanded on and more neat, thanks for the feedback
This thread makes me feel like a fucking ape
Coming back to this, can we prove it from contradiction, would this be a valid contradiction statement?
"There exists non-zero values for a, b, c such that a*sqrt(2) + b*sqrt(3) + c*sqrt(5) is rational",
we make the assumption a*sqrt(2) + b*sqrt(3) + c*sqrt(5) = r where r is non-zero and rational,
when we arrive to the contradiction we could say it must be that a = b = c = 0.
Unsure if that contradiction statement is valid or proving it along those lines. The other way I think is using is using matrices but I can't remember it, been so long since I've used them
This is correct (and even a good starting point) altho you cannot assume that r is nonzero. That said, it is just that, a starting point.
It's indeed possible to prove this utilizing matrices, altho it's somewhat overkill in this case. For the bonus question, however...
I basically said what I did because I was too lazy to write up my own answer and your answer contained the essential bit anyways, so I essentially just expanded on your answer. Thanks for reacting positively to my unasked for feedback btw.
Im actually suprised how many people are interested in maths here, I didn't expect to see so many high iq cel
I assume it's because its what we're trying to get to?
Continuing from what I wrote earlier, we square both sides to get (a*sqrt(2)+ b*sqrt(3) + c*sqrt(5))^2 = r^2.
Expanded, this is 2ab*sqrt(6) + 2ac*sqrt(10) + 2bc*sqrt(15) + 2a^2 + 3b^2 + 5c^2 + 2ab*sqrt(6) + 2ac*sqrt(10) + 2bc*sqrt(15),
simplifying we get 2ab*sqrt(6) + 2ac*sqrt(10) + 2bc*sqrt(15) + 2a^2 + 3b^2 + 5c^2 = r^2.
r^2 is rational because r is rational, this equation implies that the sum of rational numbers 2a^2, 3b^2, and 5c^2, and the sum of irrational numbers 2ab*sqrt(6), 2ac*sqrt(10), and 2bc*sqrt(15), must also be rational. We know that 2a^2, 3b^2, and 5c^2 and its sum is rational.
However, for any rational number x, the product of an irrational number y and x is always irrational, unless if y is zero. Thus, for the irrational terms 2ab*sqrt(6), 2ac*sqrt(10), and 2bc*sqrt(15) to combine to a rational sum, each of the coefficients a, b, and c must be zero. This means that a = b = c = 0, contradicting our initial assumption which was that a, b, and c are non-zero. We can simply input 0 to prove the other side of the "iff" part.
Therefore this means that a*sqrt(2) + b*sqrt(3) + c*sqrt(5) is rational if and only if a = b = c = 0.
I think using the properties of linear independence would be suffice for this problem, since you cant express it as a combination of one using rational constants it is therefore linear independent and linear independence signifies that all the variables is equal to 0, thus a = b = c = d = 0. Is the formal proof along the lines of this logic?
all good was not even sure i did the problem right
No. The statement you're trying to prove is "if a*sqrt(2) + b*sqrt(3) + c*sqrt(5) is rational, then a = b = c = 0". By way of contradiction, you therefore assume that a*sqrt(2) + b*sqrt(3) + c*sqrt(5) is rational (say r) for some rational a & b & c not all zero. It's (at this stage) not clear that our hypothetical a & b & c cannot result in r = 0.
This is unfortuantely not the right approach.
You accidently wrote every cross term twice here, altho you corrected this immediately afterwards.
This does not hold. If this logic were to suffice, then you could've applied this logic to the original a*sqrt(2) + b*sqrt(3) + c*sqrt(5) instead and saved yourself some hassle. Here's why this logic doesn't work -- by your logic, since each of a*sqrt(5) and b*φ (φ being the golden ratio) is irrational, the only way they can combine to yield a rational number is if a = b = 0; however, 2*φ - 1*sqrt(5) = 1, so this example disproves your assertion.
Brutal counter-example, how would you solve it starting from here then?
Hint: rearrange before squaring.
Square both sides of a*sqrt(2) + b*sqrt(3) = r - c*sqrt(5). If you do, you'll see that you've reduced the total number of square roots from three to two. Rearranging to get both square roots on the same side (and everything else on the other side) and squaring again will bring you down to only one square root -- a well-known case.
I think you solve this by proving that sqrt(2), sqrt(3), sqrt(5), sqrt(7) are linearly independent
This means that
There is no a,b,c ∈ ℚ such that :
sqrt(2)= a*sqrt(3)+b*sqrt(5)+c*sqrt(7)
sqrt(3)= a*sqrt(2)+b*sqrt(5)+c*sqrt(7)
sqrt(5)= a*sqrt(3)+b*sqrt(2)+c*sqrt(7)
sqrt(7)= a*sqrt(3)+b*sqrt(5)+c*sqrt(2)
We can recall that:
Rational* irational=irational (So long as the rational is no 0 so eg( 0*sqrt(3)=0, but for anything else it would be irational)
irational+irational=Can be irational or rational
sqrt(5)+-sqrt(5)=0 (Rational) [It requires you to be able to create sqrt(5) from sqrt(2),sqrt(3),sqrt(7) which is impossible for this to happen]
sqrt(5)+ sqrt(3)= Irrational
However for it to form a rational number it would have to linearly dependent, since it would need to be able to create it (If that makes sence)
This is equivalent to what is being asked, yes.
All of this is true, but I fail to see how these observations help in any way. You started and ended with linear (in)dependence, so you went in a circle without really proving anything.
A question on the philosophical side. Do you think math is invented or discovered? I personally believe that math is invented since a lot of the results we "derive" are contingent on the unquestioned assumptions baked into the language. For example, what does it even mean to "multiply" two complex numbers. In essence we are applying the distributive property of addition to the numbers. But this begs the question of what does it even mean to "add" a real number and an imaginary number to form a complex number. Because the act of multiplication is contingent on the assumption that a real number and an imaginary number can be added at all
I think it's a mix of both. Your argument is compelling -- esp. at the axiomatic level. Something like a Vitali set (a pathological subset of the real numbers which cannot reasonably be assigned a length) is surely an artefact of our constructions which is not to be discovered anywhere. Yet, on the other hand, can we really say to have invented trigonometry when it was almost surely conceived to try and describe celestial motions? Perhaps I'll put it like this -- mathematics might be an invention to unify various related discoveries we've made. I quite like the sound of that at this late hour.
Hate maths
Here's a number theory problem. Does there exist an integer x such that x^2=2 (mod 209) ?
God, it's been such a long time since I did one of these. You're basically asking whether 2 is a quadratic residue mod 209. Since 209 = 11*17, we have to use Jacobi symbols. I'll write (m|n) for the Jacobi symbol.
(2|209) = (2|11)*(2|17) = (−1)*(+1) = −1 because a supplement to quadratic reciprocity says that
- (2|p) = +1 if p is prime and ±1 mod 8
- (2|p) = −1 if p is prime and ±3 mod 8
@CountBleck if you'd be interested in solving some problems, feel free to give me an area of mathematics and I'd be more than happy to see if I know of a fun problem. Please no algebraic or differential topology tho
haha nice. probably gave you flashbacks to your first year or second year in university
do you got any cool problems about group/ring theory ?@CountBleck if you'd be interested in solving some problems, feel free to give me an area of mathematics and I'd be more than happy to see if I know of a fun problem. Please no algebraic or differential topology tho
Early third actually if I'm not mistaken. But I like quadratic reciprocity so they were nice flashbacks.
This one's fairly classic, but I like it a lot. Lemme know if you've come across it before.
Prove that any Boolean ring (a ring where x^2 = x for all x) is commutative.
Let R be a boolean ring, and let a, b be arbitrary elements in R.
Then (a+b)^2 = a^2+ab+ba+b^2 = a+ab+ba+b=a+b
Since (R,+) is a group it follows that ab+ba=0. (*)
Now if r is in R, note that
(r+r)^2=r^2+r^2+r^2+r^2=r+r+r+r=r+r, and since (R,+) is a group, it follows that r+r=0.
In particular, every element in a boolean ring is it's own additive inverse. So back to (*), if we can write
ab+ba = 0 = ab+ab
from which it immediately follows that R is commutative.
I never learned about boolean rings in any of my abstract algebra courses.
Very impressive. It was an exercise back when I did ring theory. At the time, I couldn't figure it out for the life of me. After looking it up, I was like "wow". Props you managed to figure this out.
If I may be so bold, however, while your proof was flawless, I think it'd've been slightly cleaner to write
a + b = (a + b)^2 = ... = a + ab + ba + b instead of (a + b)^2 = ... = a + ab + ba + b = a + b as you did. That way every individual equality logically follows from earlier results or axioms and the conclusion follows from comparing the left- and rightmost terms. It just flows a little bit better IMHO. Ditto for r calculation. Then again, I suspect you might've already known that, as you did write ab + ba = 0 = ab + ab in the "elegant" order.Anyhoo, very nicely done.
what is 1,009 rounded to the nearest hundred? pls help i'm retarded.
thanks sir
i'm tired of being a major brainlet
, can you recommend any textbooks that'll start me off with the very basics in math? I'm talking like from adding,subtracting,division,fractions type stuff all the way up to pre algebra?
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I've never used books like that myself, so I wouldn't know which ones to recommend. Sorry.i'm tired of being a major brainlet
C
autism: the thread
pls saar help me
, i must learn mathematic! to feed my poor curry family like Srinivasa Ramanujan
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