Gokubro
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PLEASE, post a math problem that is simple enough for me to solve to feel smart BUT not too hard that I can't solve it.
@Grim_Reaper @CountBleckLet A be an n by n matrix satisfying A^3 = 0. Prove that rank(A^2) ≤ n/2.
What's your level?PLEASE, post a math problem that is simple enough for me to solve to feel smart BUT not too hard that I can't solve it.
I'm not sure, slightly higher than an average normie who barely knows math.What's your level?
High school level math then? How about this problem:I'm not sure, slightly higher than an average normie who barely knows math.
Since A^3 = 0, then im(A^2) is a subset of ker(A^2) which is a subset of ker(A). Let k be the dimension of im(A^2) and let n-k be the dimension of ker(A^2). Then dim(im(A^2)) ≤ dim(ker(A^2)) so k ≤ n - k which implies k ≤ n/2 so rank(A^2) ≤ n/2@Grim_Reaper @CountBleck
The only way for this to be true is if A^2 = 0, which is not necessarily the case. Consider for example a 3 by 3 Jordan block with 0 along the diagonal.im(A^2) is a subset of ker(A^2)
Would it be safe to assume that rank(A^2) ≤ rank(A), so rank (A^2) + rank(A) ≤ rank(A^2*A) + n -> rank(A^2) + rank(A) ≤ n so rank(A^2) ≤ n/2?The only way for this to be true is if A^2 = 0, which is not necessarily the case. Consider for example a 3 by 3 Jordan block with 0 along the diagonal.
Yes, that'd be correct. The rank of a matrix is the dimension of its image, so the obvious im(A^2) ⊆ im(A) implies that rank(A^2) ≤ rank(A). Nicely done.Would it be safe to assume that rank(A^2) ≤ rank(A), so rank (A^2) + rank(A) ≤ rank(A^2*A) + n -> rank(A^2) + rank(A) ≤ n so rank(A^2) ≤ n/2?
If linear algebra was a girl, I'd so fuck her.Fucking linear algebra.
If lin alg was a girl, I'd hate fuck her. I loved the abstract algebra and group theory that came later, but lin alg can go to hell.If linear algebra was a girl, I'd so fuck her.
If you like group theory but hate linear algebra, how do you feel about representation theory?If lin alg was a girl, I'd hate fuck her. I loved the abstract algebra and group theory that came later, but lin alg can go to hell.
I haven't looked into it. It's covered in graduate algebra courses that I didn't take.If you like group theory but hate linear algebra, how do you feel about representation theory?
Linear algebra would be a bratty teen slut while group theory would be a roastie.If lin alg was a girl, I'd hate fuck her. I loved the abstract algebra and group theory that came later, but lin alg can go to hell.
Group theory would be the porn milf with the body of a 25 year old. Accurate representation (KEK) of lin alg, tbh.Linear algebra would be a bratty teen slut while group theory would be a roastie.
Linear algebra would be a bratty teen slut while group theory would be a roastie.
What do you base this on?Group theory would be the porn milf with the body of a 25 year old. Accurate representation (KEK) of lin alg, tbh.
Just a preference, sweaty.What do you base this on?
h = g^2 + z / 1000 * y
h = g^2 + z / 1000 * y
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.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
Let S(n) be the sum of proper divisors of n. We equivalently show that S(n) < n*ln(n) for n > 1.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
dang you're quick good job. i wish we had a LaTex embedd somehow so it woulda been a lot easier to read thoLet 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)
Thanks. Yeah, I know it reads like ass, but I was too lazy to externally LaTeX it, sorry.dang you're quick good job. i wish we had a LaTex embedd somehow so it woulda been a lot easier to read tho
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.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?
Heres a fun problem that even someone who doesn't know too much could solvePLEASE, post a math problem that is simple enough for me to solve to feel smart BUT not too hard that I can't solve it.
It works because you have set this condition of "if it's not 9 - recount it" so sooner or later it will become 9.The question is how do you prove why this process works?
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)
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.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
For those interested, this process is called taking the digital root btw.The question is how do you prove why this process works?
yea fair enough, defs could've been expanded on and more neat, thanks for the feedbackIn 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.
Let a,b,c be rational numbers. Prove that a*sqrt(2) + b*sqrt(3) + c*sqrt(5) is rational if and only if a = b = c = 0.
Bonus (much harder): Let a,b,c,d be rational numbers. Prove that a*sqrt(2) + b*sqrt(3) + c*sqrt(5) + d*sqrt(7) is rational if and only if a = b = c = d = 0.
Coming back to this, can we prove it from contradiction, would this be a valid contradiction statement?Unfortunately, I don't think so. The problem I see is with the following step:
In general, the sum of two irrational numbers can be rational (e.g., add π to 4 - π). I wouldn't know how the prove the aforementioned step (from the smaller cases). If you have an idea, however, I'd love to hear it.
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."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.
It's indeed possible to prove this utilizing matrices, altho it's somewhat overkill in this case. For the bonus question, however...The other way I think is using is using matrices but I can't remember it, been so long since I've used them
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.yea fair enough, defs could've been expanded on and more neat, thanks for the feedback
I assume it's because its what we're trying to get to?This is correct (and even a good starting point) altho you cannot assume that r is nonzero.
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?It's indeed possible to prove this utilizing matrices, altho it's somewhat overkill in this case. For the bonus question, however...
all good was not even sure i did the problem rightI 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.
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.I assume it's because its what we're trying to get to?
This is unfortuantely not the right approach.Continuing from what I wrote earlier, we square both sides to get (a*sqrt(2)+ b*sqrt(3) + c*sqrt(5))^2 = r^2.
You accidently wrote every cross term twice here, altho you corrected this immediately afterwards.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),
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.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.
Brutal counter-example, how would you solve it starting from here then?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.
"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 rational
Hint: rearrange before squaring.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 independentLet a,b,c be rational numbers. Prove that a*sqrt(2) + b*sqrt(3) + c*sqrt(5) is rational if and only if a = b = c = 0.
Bonus (much harder): Let a,b,c,d be rational numbers. Prove that a*sqrt(2) + b*sqrt(3) + c*sqrt(5) + d*sqrt(7) is rational if and only if a = b = c = d = 0.