How to Prove Stuff in Math

Here are four methods of proof:

Direct proof.

Proof by contrapositive.

Proof by induction.

Proof by contradiction.

We begin by trying to prove an IF THEN type of statement. If belief P is true, Q will also be true, P implies Q. Or, P==>Q.

A direct form of proof assumes P and proceeds directly to Q. To prove by contrapositive, one shows that NOT Q ==>NOT P by one of the methods of proof, and then takes the contrapositive of NOT Q==>NOT P to prove P===>Q.To prove by induction,form n terms of data and show that if the nth term is true, that the nth +1 term follows.To prove by contradiction, begin with P===>Q, assume Q false and P true, and show that this condition leads to a false conclusion of some type. Then P==>Q and both are really true.

Let’s try each.

First we begin by defining our terms. Suppose we wish to try and prove the following: THE SUM OF TWO ADJOINING INTEGERS IS AN ODD INTEGER.

In other words: If I and J are adjoining integers, then I + J is an odd integer.

An integer is a counting number and its inverse. We count one object at a time. The collection of an object and another object are 2 objects, etc. If m is a counting number, then m, m+1 and m-1 are integers.

I is ODD means there exists k an integer such that I=2k+1.

An adjoining integer means that if I is some integer, then the adjoining ones J equal I+1 and I-1.


CASE 1: Let I be an integer. Then I +1 is adjoining. Let S be their sum. = I + I+1=2I+1.By definition, S is odd.

CASE 2 Let I be an integer. Then I-1 is adjoining. Let S be their sum =I+I-1. S=2I-1.Let k=I-1. I=k+1. S=2(k+1)-1.S=2k+1. By definition, S is also odd.


Assume S not odd. Then S is even or S=2k where k is also an integer.

CASE 1. Let I be an integer. I+1 is also an integer. S=I+I+1. S=2I+1.If S is even and not odd, S=2k. Then 2k=2I+1.k=I+1/2.k is not a counting number.k is not an integer.But this

is is a contradiction. Therefore S is not even. Therefore S is an odd integer.


Let I be an integer. I-1 is adjoining.S=I+I-1.Assume S is not odd. S is even. Then there exists k an integer k such that 2k= I+I-1=2I–1. k= I- 1/2.Therefore k is not an integer. This is a contradiction. S not even. Therefore S odd.



Let I be an integer. Then I + 1 is an adjoining integer.

Consider sums of adjoining integers I and I+1: -2+-1=-3,-1+0=-1,0+1=1,1+2=3,2+3=5, etc. Assume the nth term is odd. We want to prove that the nth +1 term is odd.Let S be that nth sum.S= 2k+1. In the above series,S=-3 for n=1,S=-1for.n=2, S=1 for n=3, etc. So S=2k+1=2(n-1)-3=2n-5.k=n-3 an integer. S is odd The nth plus 1 term is R=2(n+1)-5=2n-3.Choose k=n-2. R=2(k+3)-5=2k+1. R representing nth +1 sum.R is odd.Therefore the sum of the adjoining integers I and I+1 is odd.


Let I and a adjoining integers. Consider the series I +I-1: 3+2=5,2+1=3,1+0=1,0+-1=-1,-1+-2=-3 etc.Let S be the sum equal to the nth term, 5,3,1,-1,-3…S is odd.S=-2n+7. We want to know if there is an integer k such that 2k+1= -2n+7.2k=-2n+6. k=-n+3. The answer is yes. The sum is odd.Let R be the sum of the nth +1 term.R=-2(n+1)+7=-2n+5=2k+1 to be odd. We want to know if integer k exists.k=-n+2. Yes. The sum of the nth+1 term is odd. The sum of two adjoining integers I and I-1 is odd.


We wish to prove that the sum of two adjoining integers is odd.To do this, we prove that if the sum of two integers is NOT odd,we show that they are NOT adjoining.

P===>Q. NOT Q===>NOT P.

S is not odd. Therefore S is even. Therefore S=2k where k is an integer.Then 2K=I+J. K= I/2 +J/2. Then K =qi+ri/2+qj+rj/2 where qi and qj are the quotients and ri and rj the remainders. The remainders are both 0 or both 1 if K an integer.If both 0,2K= qi+qj.Then qi and qj are both even or both odd.I and J are non adjoining.If both remainders are 1, I=2qi+1 is odd.J=2qj+1 is odd.I and J are non adjoining if one includes I=J as non adjoining.

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