METHODS OF PROVING :)

METHODS OF PROVING :)

1. INDUCTION

   • Prove by mathematical induction: 1/1.3 + 1/3.5 + 1/5.7 + … + 1/ (2n+1)(2n+1) = n/2n+1 If we denote that sum by S(n), then assume that the formula is true for n = k; that is, assume S(k) = k/2k + 1 . Now show that the formula is true for n = k + 1; that is, show: S(k + 1) = k + 1/2k + 3 . Begin: S(k + 1) = S(k) + The next term, whose denominator is the product of the next odd numbers. = k/2k+1 + 1 / (2k+1)(2k+3) = k(2k+3)+1 / (2k+1)(2k+3) = 2k2 + 3k+1 / (2k+1)(2k+3) = 2k2 + 3k+1 / (2k+1)(2k+3) = (2k+1)(k+1) / (2k+1)(2k+3) Next, The formula is true for n = 1: 1/1.3 = 1 / 2.1+1 = 1/3 Therefore it is true for all natural numbers.
   • If 1) when a statement is true for a natural number n = k, then it will also be true for its successor, n=k+1; and 2) the statement is true for n = 1; then the statement will be true for every natural number n.

2. CONTRADICTION

   • establishes the truth or validity of a proposition by showing that the proposition's being false would imply a contradiction
   • If a, b∈Z, then a2−4b−36=0. Proof. Suppose for the sake of contradiction that a, b∈Z but a2−4b−3=0. Then we have a2=4b+3=2(2b+1)+1, which means a2 is odd. Therefore a is odd also, so a=2c+1 for some integer c. Plugging this back into a2−4b−3=0 gives us (2c+1)2−4b−3 = 0 4c2+4c+1−4b−3 = 0 4c2+4c−4b = 2 2c2+2c−2b = 1 2(c2+c−b) = 1. From this last equation, we see that 1 is an even number, a contradiction.

3. DIRECT PROOF

   • P⇒Q
     • If p, then q.
   • Straight forward proving method for conditional statements.
   • If x and y are positive real numbers, then 2pxy≤x+y. Proof. Suppose x and y are positive real numbers. Then 0≤(x−y)2, that is, 0≤x2−2xy+y2. Adding 4xy to both sides gives 4xy≤x2+2xy+y2. Factoring yields 4xy≤(x+y)2. Previously we proved that such an inequality still holds after taking the square root of both sides; doing so produces 2pxy≤x+y

4. CONTRAPOSITIVE

   • ~P⇒~Q
     • Statements P⇒Q and ∼Q⇒∼P are different ways of expressing exactly the same thing.
   • Alternative to direct proof.
   • Suppose a,b∈Z. If a2(b2−2b) is odd, then a and b are odd. Proof. (Contrapositive) Suppose it is not the case that a and b are odd. Then, by DeMorgan’s law, at least one of a and b is even. Let us look at these cases separately. Case 1. Suppose a is even. Then a=2c for some integer c. Thus a2(b2−2b) =(2c)2(b2−2b)=2(2c2(b2−2b)), which is even. Case 2. Suppose b is even. Then b=2c for some integer c. Thus a2(b2−2b) =a2((2c)2−2(2c))=2(a2(2c2−2c)), which is even. (A third case involving a and b both even is unnecessary, for either of the two cases above cover this case.) Thus in either case a2(b2−2b) is even, so it is not odd.

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