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- Apr 22, 2007
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For those of you who love math as much as I do, here are some fun math things. Try to contribute some good problems to the thread and I'll update the OP with them if I like them. Also, post what's wrong with the problems in the OP, if anything, and I'll update it with the correct solutions (mathematical rigour is preferred, but I doubt most of you are literate in the formalities of mathematics)!
It is known that:
ln 2 = 1 - 1/2 + 1/3 - 1/4 + ...
So, consider the following:
ln 2 = 1 - 1/2 + 1/3 - 1/4 + ...
Partition the terms into odd and even parts:
ln 2 = (1 + 1/3 + 1/5 + ...) - (1/2 + 1/4 + 1/6 + ...)
= (1 + 1/3 + 1/5 + ...) + (1/2 + 1/4 + 1/6 + ...) - 2*(1/2 + 1/4 + 1/6 + ...)
Combine the first two terms and distribute the 2 into the third term:
ln 2 = (1 + 1/2 + 1/3 + ...) - (1 + 1/2 + 1/3 + ...) = 0
eln 2 = e0
2 = 1
Consider the expression: the smallest positive integer not representable in under twelve English words.
Does such an integer exist? (this problem is tricky as shit)
Contributed by Zarathustra
Which is greater, the number of natural numbers or the number of squares of natural numbers? Note that for every natural number, there is a corresponding square... but some natural numbers (like 2 or 3 or whatever) are not the squares of natural numbers.
For those of you who are total noobs, the natural numbers is the set of positive integers, sometimes including 0, i.e. {0, 1, 2, 3, 4, ...}
It is known that:
ln 2 = 1 - 1/2 + 1/3 - 1/4 + ...
So, consider the following:
ln 2 = 1 - 1/2 + 1/3 - 1/4 + ...
Partition the terms into odd and even parts:
ln 2 = (1 + 1/3 + 1/5 + ...) - (1/2 + 1/4 + 1/6 + ...)
= (1 + 1/3 + 1/5 + ...) + (1/2 + 1/4 + 1/6 + ...) - 2*(1/2 + 1/4 + 1/6 + ...)
Combine the first two terms and distribute the 2 into the third term:
ln 2 = (1 + 1/2 + 1/3 + ...) - (1 + 1/2 + 1/3 + ...) = 0
eln 2 = e0
2 = 1
Consider the expression: the smallest positive integer not representable in under twelve English words.
Does such an integer exist? (this problem is tricky as shit)
Contributed by Zarathustra
Which is greater, the number of natural numbers or the number of squares of natural numbers? Note that for every natural number, there is a corresponding square... but some natural numbers (like 2 or 3 or whatever) are not the squares of natural numbers.
For those of you who are total noobs, the natural numbers is the set of positive integers, sometimes including 0, i.e. {0, 1, 2, 3, 4, ...}
For each natural number n, associate with n the number n2. And for each square n2, associate the natural number n. Since you can associate exactly one square with one natural number, and exactly one natural number with one square, then the set of natural numbers and the set of squares of natural numbers must have the same size.
For all of you noobies, this is called a bijection. A bijection is defined as an injection and a surjection.
An injection is a one-to-one function, that is if f(x) = f(y), then x = y. This means that for every element in the domain, there is exactly one mapped element in the range.
A surjection is an onto function, that is, for any element y in f's range, there exists an element x in f's domain such that f(x) = y.
Here is a more formal solution:
Consider the bijection f: ℕ → {n2 : n ∈ ℕ}, defined as f(n) = n2
f is injective since:
f(n) = f(m) ⇔ n2 = m2 ⇔ n = m, since the natural numbers are positive.
f is surjective since for any square n2, f(n) = n2
So, f is indeed a bijection.
Since f is a bijection, then the size of f's domain and range is the same. (this is an easy result)
For all of you noobies, this is called a bijection. A bijection is defined as an injection and a surjection.
An injection is a one-to-one function, that is if f(x) = f(y), then x = y. This means that for every element in the domain, there is exactly one mapped element in the range.
A surjection is an onto function, that is, for any element y in f's range, there exists an element x in f's domain such that f(x) = y.
Here is a more formal solution:
Consider the bijection f: ℕ → {n2 : n ∈ ℕ}, defined as f(n) = n2
f is injective since:
f(n) = f(m) ⇔ n2 = m2 ⇔ n = m, since the natural numbers are positive.
f is surjective since for any square n2, f(n) = n2
So, f is indeed a bijection.
Since f is a bijection, then the size of f's domain and range is the same. (this is an easy result)
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