>>1 is a retard.
>>n+1 is a retard.
Therefore, ITT retardity.
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Anonymous2007-05-03 14:18 ID:MReE7WyH
I am kidding. This is actually a potentially interesting thread.
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Anonymous2007-05-03 16:24 ID:VrhrNA41
Claim: For every positive integer n, n is equal to every positive integer less than n.
Proof: For n = 1, there are no positive integers less than 1 so the claim holds trivially. Suppose the claim holds for some fixed n, i.e. that n = n - 1 = ... = 1. Adding 1 to the leftmost equality from this, n = n - 1, it follows that n + 1 = n, so n + 1 = n = n - 1 = ... = 1. Therefore the claim holds for n + 1. QED
There is only one positive integer.
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Anonymous2007-05-03 16:42 ID:U6G4vPKt
>>4
Haha, almost, except that the proof doesn't progress from 1 to 2, so you'd need to check truth for n = 2, which is almost like dividing by zero.
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Anonymous2007-05-04 1:26 ID:QFL9digi
prove by induction that
|cosx -sinx|^n = |cosnx -sinnx|
|sinx cosx| |sinnx cosnx|
where the bars are my attempt to represent a matrix.
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Anonymous2007-05-04 2:08 ID:++os15jl
rotating a point by x degrees n times is equivalent to
rotating a point by nx degrees 1 time
Proof: Consider a set of humans with 1 person in it. He is trivially the same height as himself.
Assume true for all sets of humans with < n people inside.
Consider a set of n people. Create 2 subets, each with n-1 people inside. By assumption, all the people within the subsets are the same height. Therefore, the n people are all of the same height.
Hypothesis: High voltage can be bad for you.
You can prove this taking one end of a coil of copper wire in each hand and standing in a place with a strongly fluctuating magnetic field.
Thus; proof by induction.
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Anonymous2007-05-09 17:53 ID:L9eumvUR
Use induction to solve (((n! / n^m) <= 1) for all (m >= 1))
>>27.
Shit, I mean m^n, and most likely n >= 1, not m. But then well fuck I forgot half the problem already.
Anyway, the proof involves expanding n! into (n)(n - 1)(n - 2)...(2)(1), and expanding m^n into (m)(m)(m)(m)...(m) n times. Since it's a quotient, you have to show that the whole thing expands into (n / m) * ((n-1)/m) * ((n - 2)/m) *... (1/m) for n terms.
Well, this would have been clever if I had remembered exactly what I was talking about before I typed it all out. Sorry.
>>26,28
almost, but nice try; what you're trying to say is a good example of induction
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Anonymous2007-05-10 8:50 ID:0BDXdN6r
/r/ INDUCTION PLZ
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Anonymous2007-05-10 9:22 ID:cbjS+xta
Lemma 8. On the Cohen poset P, any set of incompatible conditions is necessarily countable.
(One also says that P satisfies the "countable chain condition" because it means that any "antichain" in P is countable!)
Proof: Given a set A of incompatible conditions, let An be the set of those conditions p in A for which the domain F_p of p consists of exactly n elements. Then it clearly suffices to show that each A_n is a countable set. This will be done by induction on n. For n = 0 there is nothing to prove. So suppose we have proved that any set of mutually incompatible conditions, each of n - 1 elements, is a countable set. To prove that A_n must thus be countable also, write A_n = Union_m A_{n,m}, where
A_{n,m} = {p in A_n | Exists b in B such that p(b,m) is defined}.
It is then enough to show that each A_{n,m} is countable. Pick for each p in A_{n,m} a b_p in B such that p(b_p, m) is defined, and write, for i = 0,1,
A_{n,m,i} - {p in A_n | p(b_p,m) =i}.
Since for any m and i, the elements of A_{n,m,i} are pairwise incompatible, so is the set of their restrictions, written p|...,
R_{n,m,i} = {p|F_p- {(b_p, m)} : p in A_{n,m,i} }.
This is a set of conditions on n - 1 elements, hence a countable set, by the induction hypothesis. Therefore A_{n,m,i} is countable, and hence so is A_{n,m} = A_{n,m,0) union A_{n,m,1} This completes the proof.
(MacLane + Moerdijk, Sheaves in Geometry and Logic, page 290)