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#177 Re: Jai Ganesh's Puzzles » 10 second questions » 2010-02-09 02:10:19
#2025
It is well known that
Therefore
Hence
Thats how you do it in 10 seconds!
#178 Re: Exercises » Is this cool with you? » 2010-01-09 05:32:17
She manipulates this to:
She says proved:
The question was, is her manipulation a proof. Do you like where she stops? Do you find her proof valid? Is more required? When does algebraic manipulation constitute a proof?
So you just want to prove
for all natural numbers n (even though this is the wrong way to go about solving the original question of finding a square between n and 2n)?
I dont know what the person did, but if we want to use her method, we should proceed as follows. Start by noting that for all integers n ≥ 1,
Hence for all natural numbers n,
and the proof is completed by adding n² throughout.
The wrong way to go about it is to start with [1] and get [2]. This would be assuming what you want to prove to be true already, and would therefore prove nothing yet this is an all too common mistake among students (and sometimes even teachers).
So what did the person do? Did she start from [1] and get [2] (which would be incorrect) or did she start with [2] and get [1]?
#179 Re: Exercises » Is this cool with you? » 2010-01-08 23:17:48
Hi;
How would you judge this answer? And why?
The first step is invalid because in order to split the limit of a sum as the sum of limits, you have to make sure all the limits you are taking exist. In this case, none of the three limits exists. The rest of the proof is treating ∞ as a number, which it is not.
The second problem looks fine to me.
The third problem can also be proved this way. Note that
can be written
since n is positive. There are n+1 integers from n to 2n inclusive and so there are at least n+1 integers between k² and (k+1)². Hence these two integers must differ by at least n+2:
Combining with k² < n gives
which is another contradiction.
#180 Re: Exercises » The Golden Ratio » 2009-11-22 17:34:57
If ϕ is the golden ratio, ϕ² should be ϕ+1, not ϕ−1.
#181 Re: Help Me ! » Algebra problem » 2009-11-14 07:18:23
I'm glad I was able to help! Now that you've done your homework, you might like to check your solutions with mine.
#182 Re: Help Me ! » Algebra problem » 2009-11-13 11:21:55
if anybody knows this, pls help...
1. determine a group of Aut((Z/2Z) ⊕ (Z/2Z)) (prove every statement)
2.show that cycle (of length n) is even permutation if and only if n is an odd number
3. G={a1, a2, ..., an } finite abelian group. Is it always (a1 a2 .... an)^2=e ?
1. This is the automorphism group of the Klein 4-group; I think it is but you will need to verify it.
2. Hint: A cycle of length n can be written as a product of n−1 transpositions.
3. Hint: Separate those elements which are their own inverses and those which are not; for the latter, pair up the elements with their inverses.
#183 Re: Puzzles and Games » A test problem... » 2009-11-13 11:01:30
none of dont have .
To generalize:
Suppose there are teachers and given , we want that any or more teachers know all the answers but no or fewer teachers will know all the answers. Then the minimum number of questions is and each teacher is given answers.
That is a very good generalization. 