# Proof by Deduction

In this article, we will learn “how to prove a statement by deduction?”

## Content

• Introduction
• Definition
• Methodology
• Practical Examples
• Summary

## Introduction

In mathematics proving some statement is always a critical work. Because any statement in mathematics has its significance. Here we are discussing one of the techniques to prove whether the statement is true or false? A mathematical deduction is a well-known method to prove any statement in mathematics. In this method, we make some logics to prove the statement true.

In this technique of proof by deduction, we first consider the statement is true for all cases. Word “Deduction” means to find out the conclusion by making some observations. Based on observation, we deduce the results. The whole procedure is done through by well-known mathematical principles.

“Proof by deduction” is a very important technique in mathematical science. After proving any statement through this method is always considered to be true for every case. There are also many techniques to prove the mathematical statement, but proof by deduction has its extraordinaryvalue.

In mathematics proving any statement is an art. When we have to prove any statement universally, we use “proof by deduction” whenever a statement has to show, that we indeed say it is conjecture, then we make some logic to prove it, if it proved then it is said to be a theorem.

## Definition

Proving a mathematical statement by making logic and using mathematical principles is called “proof by deduction”.

What logic we have to made or what principles we have to use? We will see them in the section on methodology.

## Methodology

To prove any mathematical statement by deduction, we first consider some mathematical concepts that are called “axioms”. Through these axioms, we made some logic and deduce the result.

Proof by deduction based on logic, secondly make some logic and start work.

For example, we have to prove the given statement.

Jenny is a girl, so she loves Barbie dolls.

Here are two parts in the statement, one is “Jenny is a girl” and the second one is “she loves Barbie dolls”. Consider that the first statement is A, and the second statement is B. It means If “A” then “B”, this is our logic for the statement. Mathematically we will write as

A⇒B

“⇒” is the symbol of “implies”, above statement will be read as “A implies B”.

Here we consider an axiom, that every girl loves Barbie dolls. Through this axiom, we conclude our results as If Jenny is a girl, then she must love Barbie dolls.

If Jenny is a girl, and she doesn’t love the Barbie dolls based on our logic then our statement becomes incorrect.

For using the technique “proof by deduction” our mathematical axioms and logic should be strong. If at any stage we use wrong axioms or logic, it yields us the wrong result.

Here is an example,

Consider x = y

Then, we can write as,

$x^{2}=y^{2}$

Because x = y, their square must be equal,

$x^{2}=yy$

Replacing one y with x, because x=y

$x^{2}=xy$

Subtracting $y^{2}$ on both side of the equation,

$x^{2}-y^{2}=xy-y^{2}$

Applying the formula $a^{2}-b^{2}=(a-b)(a+b)$ on the left-hand side and taking y common from the right-hand side,

x – yx + y = y x – y

Dividing by (x – y) into both sides,

x + y = y

Replacing  x with y, because x = y

y + y = y

Adding both y on the left side,

2 y = y

Dividing by y on both sides

2 = 1

Is it true? Of course not, we know very well it is impossible, then where we make mistake?

We make mistake when we divided by (x – y).

Since x = y, which means x – y = 0. It’s mean in the following step we divided by 0.

x – yx + y= y x – y

Dividing by (x – y) into both sides,

x + y = y

And dividing by zero is not allowed in mathematics.

In this example, we see that at any stage we wrongly use the logic and got wrong results.

So, for proving any mathematical statement our logic should be strong.

In each statement, we follow the following steps

Make conjecture (Conjecture mean opinion)

Consider axiom.

Apply logic.

Deduce results.

We will see in the examples, how to do step by step.

Always remember if at any stage any axioms or logic is not satisfied, then it means our statement is wrong.

## Examples

Problem 1: The sum of every two odd integers is always even.

Here is the statement we have to prove by deduction.

According to our steps first, make a conjecture, that is the addition of odd numbers.

Now, consider axioms.

1. Commutative property which is a + b = b + a
2. Associative property which is a + (b + c)=(a + b)+ c
2. Odd number, if n is the integer then 2n must be even and 2n + 1 will be odd. Always 2n is even and 2n + 1 is odd for every integer n.

Now consider M and N are two odd numbers, then M = 2a + 1 and N = 2b + 1, where a and b are the integers.

Now it’s time to deduce results.

M + N = 2a + 1 + 2b + 1

M + N = 2a + 2b + 2

M + N = 2 a + b + 1

Since a , b, and 1 is the integers, the sum of these also is the integer, say the sum of these three integers is k, where k is also an integer.

So,

M + N = 2 k

As we know 2k is an even integer. So, our statement is true, that sum of any two integers always even.

To verify take any two-odd integer, let 3 and 5 are two odd integers, then.

3 + 5 = 8

Since 8 is even.

Problem 2: Prove that if a divides b, and b divides c, then a divides c.

Here we have to prove that a divides c.

If any number n divides m then, we can write m = p n , where p is an integer.

It’s mean if a divides b, and b divides c, then we can write the following.

b = k a,      and      c = l b

Since

c = l b

Replacing the value of b

c = l (k a)

c = ( l k ) a

Here l and k are the integers, their product must be an integer, say lk = m, above equation reduce to

c = m a

Which shows that a divides c.

Hence our stamen is true.

To verify let the values of a, b, and c.

Consider a = 2, b = 4, c = 8.

Since 2 divides 4, and 4 divides 8 and also 2 divides 8.

Problem 3: Prove that $x^{2}-6x+11$ is always positive.

Before going forward, we consider conjecture, that is $x^{2}-6x+11$ is greater than zero for all values of x. Here we consider axiom, which is square of each number is always positive and second is the sum of two positive number is always positive. Then we rearrange the given expression to convert it into a form where we can apply our axioms.

To solve this statement, we have to rearrange the given statement.

We will show that,

$x^{2}-6x+11>0$

Rearrange the statement as

$x^{2}-6x+9+2$

Here $x^{2}-6x+9$ can be written as $(x-3)^{2}$, so the above equation will be reduced,

$(x-3)^{2}+2$

Here $(x-3)^{2}$ is a positive number for every value of x, due to square on it.

Now $(x-3)^{2}+2$ is also positive because the sum of two positive numbers is always positive.

So

$x^{2}-6x + 11 =(x-3)^{2}+2$

Since

$(x-3)^{2}+2>0$

This implies that,

$x^{2}- 6 x + 11 >0$

Hence our statement is proved, that $x^{2}- 6 x + 11$ is positive for all values of x.

To verify consider the value of x is 5, then,

$x^{2}- 6 x + 11$

$=5^{2}-6 (5) + 11$

= 25 – 30 + 11

= 6

Since 6 is a positive number.

Problem 4: Show that $n^{2}+n$, is even for every value of n, where n is an integer.

To prove this statement, we use the following axioms.

1. As we know after every odd integer there is an even integer
2. Every two consecutive numbers have a difference of 1.
3. If n is an even, then n + 1 must be odd.
4. If n is odd, then n + 1 must be even.
5. The product of every even number with an integer is always even.

The given statement is,

$n^{2}+n$

Rewrite the above expression as:

n(n + 1)

So, if n is even, then n + 1 is odd according to our axiom and if n is odd, then n + 1 is even according to our axiom.

It means in the product of n and n + 1, one is even, and the other is odd. This means the product of n(n + 1) is the product of an even number with any other number.

So, n(n + 1) is even for every value of n.

Hence, our statement is proved that $n^{2}+n$ is even.

Problem 5: Prove that sum of every three consecutive integers is multiple of three.

Before going to prove this statement, we must consider some axioms mentioned below.

1. Every two consecutive have the difference of 1.
2. If n is an integer, then n + 1 is just an upcoming integer.
3. Similarly, n + 2 is just after the n + 1.

So, three consecutive integers are n, n + 1, n + 2.

Here we have to check that sum of three consecutive numbers. So, first, we calculate the sum of numbers.

Sum up these consecutive numbers is,

n + (n + 1) + (n + 2)

= 3n + 3

= 3( n + 1)

Since as we can see that the resulting number is 3(n + 1) since the resulting number have a factor equal to 3 it’s mean it is divisible by 3.

We can verify our results, consider the value of n. Say n = 3, then n + 1 = 3 + 1 = 4 and n + 2 = 3 + 2 = 5.

n + (n + 1) + (n + 2) = 3 + 4 + 5 = 12

The sum is 12, which is divisible by 3.

Problem 6: If (x – a) is the factor of f(x), then show that f(a) = 0.

Here according to a statement (x – a) is the factor of f(x), its mean (x – a) divides f(x)

So, f(x) can be written as,

f(x) = (x – a) g(x)

Where g(x) is any other function. Put x = a in the above equation,

f(a) = (x – a) g(a)

f(a) = 0 × g(a)

f(a) = 0

Hence our statement is proved.

## Summary:

In this topic, we learn the technique “Proof by Deduction”. We also see that how we go through the procedure and how we apply the logic. We also see that if we make mistake at any stage this leads us to the wrong result.

The main axioms should be clear and strong. Like how to write the consecutive number (i.e., n + 1, n + 2), how to write even and odd numbers (i.e., 2n, 2n+1 respectively), etc.

Source:

1. AS Level Mathematics Textbook by Aaran Karia
2. Pure mathematics 1 by Sophie Goldie