Proof by Contradiction: 3n + 2 is Even, N is Even

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• Simple Proof By Contradiction

  This completes the if and only if proof. We prove by contradiction that 3n+2 is even if and only if n is even.

• Integer Proofs by Contradiction

  Use the proof by contradiction method to prove: (a) If 7n + 4 is an even integer, then n is an even integer. Proof. We prove it by contradiction. Suppose that n is NOT an even integer. Then n is odd. So, n=2k+1 for some integer k.

• Discrete Math Problems: Set Operations, Counter Examples

  Counter example: if n=-1/2, then n/(n+1)=-1 which is an integer, but n=-1/2 is not an integer. 3 (a) Proof: case 1: if n is even, then n2 is even and 3n is also even, so n2+3n is even.

• Prove that a planar graph of order n>=3 and size m is maximal planar if and only if m=3n-6

  By applying Euler formula n-m+r=2, we get n-m+2m/3=2 i.e., n-m/3=2 So, m=3n-6 (2) If G is a planar graph of order n>=3 and size m=3n-6 , then G is a maximal planar graph. proof.

• Denumerability and induction

  For n+1, we have (n+1)^3-(n+1)=n^3+3n^2+3n+1-n-1=(n^3-n)+3n(n+1) From the induction assumption, we know that 6 divides n^3-n. Since n and n+1 are two consecutive integers, then one of them is odd and one of them is even. So 2 divides n(n+1).

• Geometry : Probability that Three Points on a Circle will form a Right-Triangle

  Thus the probability is P=k(n-2)/[n(n-1)(n-2)/6] =6k/[n(n-1)] (since n=2k and thus 6k=3n) =3n/[n(n-1)] =3/(n-1) So the final result is: when n is even, the probability is 3/(n-1); when n is odd, the probability is 0.

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  98111 Methods of proof Prove that the square of an even number is an even number using: a) a direct proof b) an indirect proof c) proof by contradiction Statement: the square of an even number is an even number Methods of proof a) a direct

• divergent sequence

  Proof: For each real number M, without the loss of generality, we only consider a very small M<-1 We can set N = |M| = -M, then for all n>N, we have x_n = n - 3n^2 = n(1 - 3n) = -n(3n - 1) < -N(3N - 1) < -N (because 3N - 1

• Argue why every odd order subgroup has to be a subgroup of the cyclic odd order subgroup K of index 2, and by the divisibility argument it still could be a subgroup of G and not to be contained entirely in K.

  This is a contradiction. Therefore, t must be even. So H has an element h*g^n with even odd. But the order of H is odd, we also get a contradiction. Thus this case is impossible. Therefore, all subgroups with odd order must be in K.