Man this thread reminded me of how fucking much I hate functions
sex
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mfw the last few posts in the math thread haven't been math 
We should start posting israel vs Palestine here so we can make this thread more relevantmfw the last few posts in the math thread haven't been math
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Then I guess I'll take it upon myself to save the math thread from such base brutality.We should start posting israel vs Palestine here so we can make this thread more relevant
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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.
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.
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(heigthpill + dickpill)*racepill = dogpill
Feels like I am reading this wrong. Because the condition seems to be false. Just substitute 25 for a, 125 for b and 32 for c
No, that would only work if it was sqrt(a) + root(b,3) + root(c,5) was the task.
You might have learned the proof that sqrt(2) is not rational. I guess this is essentially a generalization of that, but I'm too lazy to actually try
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Yep. To use the language of linear algebra, it's indeed well-known that {1, sqrt(2)} is linearly independent over the rational numbers. I'm essentially positing that {1, sqrt(2), sqrt(3), sqrt(5)} (even with sqrt(7) adjoined in the bonus) is linearly independent over the rationals as well.
can you say that since
is rational iff a = b = 0, ik i have to prove the smaller iff statements separately but not bothered to write it out rn
and
is rational iff b = c = 0
then the sum -
is also rational iff
a = b = c = 0
Then prove the converse statement by substituting 0 into
and finding 0 is rational
is this a legit way to do it?
Last edited:
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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.
T
This reminds me of when i used to send my homework into facebook groups in highschool and the people on there did if for free and for no reason
They just wanted to brag about knowing the subject
T
i wish that worked with uni assignments :,)
If we have a set of cases where it claims the foid enjoyed getting raped, it would be an uncountable set.
Prove that a topology on the set where foids claimed they didn't enjoy getting raped doesn't exist?
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In ZF every set admits a topology.
I have always been terrible in maths
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Let A be an n by n matrix satisfying A^3 = 0. Prove that rank(A^2) ≤ n/2.
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The smallest case is where a + b in Q implies at least a or b is in P. The trivial case, of course, is where a or b is 0. For all a + b in P where a or b > 0 and a or b in Q, a or b must be zero or the additive inverse.
Something something, proof by contradiction or exercise left for the reader. □
I should start posting my mathematical
discoveries here
discoveries here
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Quod Erat Demonstrandum? More like Qat Erat Demonstrandum.
h = g^2 + z / 1000 * y
Averagw IQ trait
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.
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What's your level?
I'm not sure, slightly higher than an average normie who barely knows math.
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High school level math then? How about this problem:
Find the three roots of 4x^3 - 3x = 1/sqrt(2). Hint: try substituting x = cos(q) and use trig identities.
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
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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.
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Fucking linear algebra.
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?
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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.
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If linear algebra was a girl, I'd so fuck her.
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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.
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If you like group theory but hate linear algebra, how do you feel about representation theory?
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I still think "clopen" is one of the best mathematical terms ever.
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I haven't looked into it. It's covered in graduate algebra courses that I didn't take.
Linear algebra would be a bratty teen slut while group theory would be a roastie.
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Group theory would be the porn milf with the body of a 25 year old. Accurate representation (KEK) of lin alg, tbh.
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