### Solutions to Proofs

Solutions to Proofs:

Prove that between any two real numbers there exists a rational number. Let a,b ∈ R such that a < b. Subtracting a from both sides gives 0 < (b−a). By the Archimedian Principle, we know that for any real numbers x and y there exists a natural number n such that n*x > y. Rearranging this, we know that x/y>1/n for any x and y. Let b−a=x/y. Thus we know that there exists some positive natural number n such that 0 < 1/n< b − a. Rearranging n this inequality, we get a < b − 1/n.

Furthermore, let m be a natural number such that m/n<b and b <(m+1)/n. Therefore a<b−1/n<m/n<b. Therefore,there exists m/n such that a<m/n<b. Thus, there exists a rational number between any two real numbers.

Proof 2: As an extreme case let us assume that we have countably many disjoint sets that are each countable.

We list our countable sets as A_1, A_2, A_3…, A_n,… for all n in natural numbers. The elements can be denoted as a_{1,1}, a_{1,2}, a_{1,3}…., a_{1,n},…, a_{2,1} , a_{2,2}…, a_{2,n}…, a_{j, 1}, …a_{j,n}…. For every A_j there exists f_j : A_j –>N. Thus this forms a matrix that is in 1:1 correspondence with NxN. This 1:1 correspondence is f(a_{i,j}) = (i,j). To show that the union of countably many countable sets is countable, we must show that there is a 1:1 correspondence between NxN and N. It is sufficient to show that there exists a 1:1 function from NxN–>N to show that NxN is countable. Let this 1:1 function be g((i,j))=2^i *3^j. This function is clearly 1:1. Thus we have established that NxN is countable. Therefore, the union of countably many countable sets is also countable.

Written

on April 1, 2014