Prove the following: For any integer n, if n^2 is even then n is even. Complete

Question

Prove the following: For any integer n, if n^2 is even then n is even. Complete the following statement then prove it: If m^2 is an even integer which is a perfect square, then the remainder when m^2 is divided by 4 is: Complete the following proof: Proposition: If a, b and c are integers for which a^2 + b^2 = c^2, then at least one of a and b is even. Proof: Suppose instead that both a and b are odd... Consider the following statement P: P: For any prime integer p, if p can be written as p = 3k + 1 for some integer k, then there exists an integer l for which p = 6l + 1. State the converse of P. State the contrapositive of P. Use a direct proof (i.e. not a proof by contradiction) to show that if A is the set A = {n Z|n = 3s + 1 for some integer s}; B is the set B = {n Z|n = 6t + 1 for some integer t}; and C is the set C = {n Z|n = 6u + 4 for some integer u}, then A = B C. 6d. Prove or disprove P. Let A_i = (-i, 1/i) for i N. Compute _i=1^infinity A_i and _i=1^infinity A_i. Give a sketch of the proof that your answers are correct. (Youi don't need to include all details, but if you use the Archimedean principle or its corollary as discussed in class, you should point out where principle is required.)

Explanation / Answer

In set A,

If s is odd, then s can be written as s=2m+1 (for some integer m)

Then n=3(2m+1)+1=6m+4 (its belongs to set C)

If s is even, then s can be written as s=2q (for some integer q)

Then n=3(2q)+1=6q+1 (its belongs to set B)

Hence, for every value of n in A either it belongs to set B or set C

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