Disproof by Counterexample

In this article, we will learn “How to disprove a statement by counterexample?”

Introduction

Some time in mathematical science we find a statement that looks true but in actual it is not true due to some reason. We need just an example to show that, the given statement is not true is called disproof by counter-example.

Disproof by counterexample is opposite procedure than proving the statement is true. Whenever we try to prove a statement is true, we look generally to and develop logic to show it true. But whenever we are going to disproof any statement, we need just a single example, which shows that the given example is not true. For example, if we say “every boy loves pizza”, we do not need to ask every boy, do you like pizza or not? We just find out a boy who did not like the pizza, and this will be our counterexample.

Disproving any statement has the same importance as proof of a statement has. In the technique of disproving a statement, we look for a single example, that’s the example called a counterexample.

Whenever any conjecture looks false, then we have to find out the counterexample to prove it wrong. For example, if we say, “all odd numbers are also prime”. This statement looks false, at that moment we give a single example to show that it is not true. To show that the statement is wrong we give a number, let’s 9 is odd but not prime. This single example disprove the statement. Now we can say  “all odd numbers are not prime”. Whenever we feel that the given statement is false, we look for a single example, which is easy rather than showing that it is false or not by logic.

Let another example to understand the value of the technique. That is if someone says “square of an integer is always equal to its two times” this conjecture looks false. How is it possible? As we know the square of a number equals the product of that number by itself. So, we can give him an example, let the integer is 3. The square of 3 is 9 and two times of 3 is 6, as you know 6 and 9 are not equals. So here is our counter-example, which disproves the statement that, “square of an integer is always equal to its two times”. If a person observs this statement for 2. Since the square of 2 is 4 and two times of 2 is also 4. By seeking this example, he makes the statement. But this statement is not true generally.

Most of the time mathematicians spend on making conjecture and then try to prove and disprove that conjecture. Whenever a statement looks true, we use proof by deduction and when looks false we search out a counterexample to show that the statement is not true.

The advantage of this technique is “we can make a new statement on limitation of numbers”. As a person says that all prime numbers are odd, but we know this statement is not true, because 2 is prime but not odd. So, by this counterexample, we can make the new statement that is “all prime numbers are odd except 2”. With the help of this technique, we can make limitation on our statement.

Definition

Disproving a statement by a single example is called disproof by counter-example. Due to which example we disprove our statement is called a counterexample.

Methodology

In this technique, we just look for a single example or a single number on basis of that example we can prove that the given statement is not true. Sometimes to find out that example, we have to check many cases. Maybe, the given statement is true for just one value, and on the other hand, the whole statement may be false for just one value.

During the procedure of disproof by counter-example, we first illustrate the statement, then start putting assumptions to find out the reason, why is the statement not correct?

Always follow the following steps to disprove any statement.

First, conclude some results based on the given statement.

Demonstrate the results.

Find out the counterexample based on the results.

Let us do in the examples.

Practical Examples

Problem 1: Disproof by counterexample, that x^{2}-x+5 is prime for every x, where x belongs to a set of integers.

Here we are to disprove the given equation, it means at least one value of x, does not satisfy the statement. First of all, understand the statement.

According to the statement, x^{2}-x+5 gives the prime values whenever we put any value of x.

Here we have to check on different values, getting much information is best to state results.

Let us start from x = -2,

Put x=-2, in x^{2}-x+5, we get,

(-2)^{2}-(-2)+5

4+2+5

11

Since 11 is prime, this is not a counterexample.

Now check for more values, let x = 2

(2)^{2}-2+5

4-2+5

7

7 is also prime. But we know the statement is false.

How do we know?

See the expression carefully,

x^{2}-x+5

As we know the square of any number cannot be prime. Now the second term is “-x” and the third term is “5”, somehow if the third term and the fourth term vanish out then just one term will be left, that is x^{2}, which can never be prime.

So, -x + 5, can vanish out, if we put x = 5. Let us see,

Put x = 5

5^{2}-5+5

25

And 25 is not a prime. So, we get a counterexample. Hence the statement is disproved by a counterexample.

Here is an important thing to be remembered, as we see the statement in this problem is true for a vast range of numbers, but we find out that there is only one number which may make this equation false.

So, disproving any statement is not so difficult. We just have to think when this not happen, always suppose the contrary.

Problem 2: Prove or disprove that if a > b, then a^{2}>b^{2} for all values of a and b, a,b\epsilon\mathbb{Z}.

Firstly, understand the statement. That is, if a number is greater than any number, then its square should also be greater. If we see the statement, it looks true. Because if one is greater than the other, then its square will also be greater, because the square is the product of this number by itself.

Now derive some results for different values of a and b.

Let a = 5 and b = 2, since 5 > 2 satisfying our “if” condition.

Now evaluate the statement

a^{2}>b^{2}

5^{2}>2^{2}

25 > 4

Which is true. 25 is greater than 4. But I know the statement is false. Why?

Just recall the definition of a square. What was that?

The Square of each number is always positive. It means if we take the square of a negative number, it gives the positive value.

As we know negative numbers are always less than positive numbers. So If we take the square of negative numbers their results should be positive and also we know in integer each negative number with the greatest value is always smaller than the positive number with the least value.

To understand this phrase, consider the values of a and b again.

Let a = 2 and b = -5. Since 2 > -5 satisfying our “if” condition.

Now illustrate the result,

a^{2}>b^{2}

(2)^{2}>(-5)^{2}

Which gives,

4 > 25

But this is not true. Because 4 is less than 25. Hence the statement is disproved.

We find a counterexample. It means the given statement is false. Its only work is to make a condition which is a > b then a^{2}>b^{2}, if and only if a, b ≥ 0.

Problem 3: Prove or disprove that all lines in xy-plane are perpendicular.

To prove or disprove this statement, firstly we have to understand the statement.

This statement is describing that all lines in the xy-plane are perpendicular to each other. Here recalls your concept of the line.

In mathematics, the line is represented by the fowling expression

y = mx + c

Where m is the slope and c is the y-intercept.

Now recall your concept about perpendicular lines. We learn a condition about perpendicular lines in lower standards, that is “lines will be perpendicular if the product of their slopes equal to -1”.

Now consider the two lines, with different slopes and intercepts.

y=m_{1}x+c_{1}

y=m_{2}x+c_{2}

Where m_{1} and m_{2} are representing the slopes. By the definition of perpendicular lines, the product of slopes should be -1.

Its mean, the given lines will be perpendicular if,

m_{1}m_{2}=-1

Here we can express easily that the statement is not true. Because there are hundreds of examples, in which the product of two numbers is not -1.

Let us consider the slope of one line is 3 and the slope of the other line is 5.

It means, we can write it as,

m_{1}=3

And

m_{2}=5

From these values of slopes, it is clear that the product of slopes is not -1.

So, we can consider a specific example, let two lines with the different slopes,

y = 3x + 1

y = 5x + 2

Since both are the lines, in xy-plane but these are not perpendicular.

Hence our statement is disproved.

Problem 4: “If m is an integer and m^{2} is divisible by 4, then m is divisible by 4”. Prove or disprove this claim.

According to the given statement if the square of a number is divisible by 4, then it is also divisible by 4.

By looking once, it looks true. But it’s not. Now we have to find out a counterexample that shows that the statement is false. For this, we first consider some perfect squares that are divisible by 4.

Here we can consider 4, 16, 36…

As these are the perfect square, can be written as

m^{2}=4

From this situation, we can evaluate the value of m. Taking square root on both sides we get,

m = 2   or     m = -2

As it is clear 2 or -2 both of them are not divisible by 4.

So here is the counterexample. Hence our statement is false.

Problem 5: if m is prime, then m + 2 also be prime, m > 2. Prove or disprove.

This statement states that the sum of every prime number is greater than 2 while 2 is prime.

First, consider some cases for m > 2.

Let some prime numbers, greater than 2 are 3, 5, 7, 11.

Now apply the given statement, which is m + 2

For m = 3

m + 2 = 3 + 2 = 5

For m = 5

m + 2 = 5 + 2 = 7

For m = 7

m + 2 = 7 + 2 = 9

For m = 11

m + 2 = 11 + 2 = 13

In these few cases, we obtain 5, 7, 9, 13. In the resulting number, there is 9 which is not prime. So, this is our counterexample to disprove the statement.

Summary

In this section, we learn how to disprove any statement by counterexample. After viewing this article, we must learn that “it is not compulsory for conjecture to be true for all values”. Commonly mathematicians used this method to find out the limitation of physical phenomena. For disproving any statement our concepts must be clear.

Source:

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