# Use a proof by contradiction to show that there is no rational number r for which r^{3} + r + 1 = 0

Number system is a way to represent numbers. In mathematics, there are different kinds of numbers present. Some of them are real numbers, Integers, whole numbers, rational and irrational numbers, even, odd, etc. Number system provides a unique representation for every number. There are also different types of number systems present in mathematics like decimal number systems, binary number systems, octal number systems, and hexadecimal number systems. So it defines a system to represent the numbers.

The value of each digit in a number is determined by three conditions.

- Face value of each digit.
- Position of the digit in a number. (Place Value)
- Base of the number system.

### Rational Number

A rational number is a real number of the form a/b where a, and b are integers with no common factor and b≠0. So any fraction with a denominator not equal to zero is considered a rational number. The difference between a fraction and rational numbers is fractions are made up of whole numbers and rational numbers are made up of Integers. Rational numbers are finite, repeating decimals. Examples of rational numbers are 1/2, -3/7, 6/5, etc.

- 3 is also a rational number because it can be expressed as 3/1.
- √4 is also a rational number because on simplification it gives 2 which is expressed as 2/1.
- 1.5 is also a rational number because it can be expressed as 3/2 (fractional form)
- Recurring decimals also come under rational numbers like 3.33333…

### Irrational Number

Irrational numbers are real numbers that cannot be represented as a simple fraction. It is a contradiction to irrational numbers because they cannot be represented in fraction form. Any real number that cannot be expressed in the form of p/q, where p and q are integers and q ≠ 0 are considered irrational numbers. Irrational numbers are infinite, non-repeating decimals. Examples of irrational numbers are √2, √5, √11, √3, etc.

### Use a proof by contradiction to show that there is no rational number r for r^{3} +r+1 = 0

**Proof:**

Given equation r

^{3}+r+1=0Let’s assume a rational number r=a/b where a, b are integers and b≠0.

There are three considerations/cases here for a, b.

Case 1: If a is an even number then b is an odd.

Case 2: If a is an odd number then b is an even number.

Case 3: Both a and b are odd numbers which don’t have any common factor i.e. 5/7Substitute r=a/b in the given equation.

r

^{3}+r+1=0 ⇒ (a/b)^{3}+(a/b)+1=0(a

^{3}/b^{3})+(a/b)+1=0(a

^{3}+ab^{2}+b^{3})/b^{3}=0Multiply the above equation with b

^{3}.a

^{3}+ab^{2}+b^{3}=0 ——->(1)

Case 1:If a is an even number and b is an odd then

a

^{3}is an even term, ab^{2}is an even and b^{3}is an odd term.So there by the sum of a

^{3}, ab^{2}, b^{3}i.e. Solving equation (1) gives an odd term but zero is an even number. So there is a contradiction here.

Case 2:Let’s move into another case where a is an odd number then b is an even number

Then a

^{3}is an odd term, ab^{2}is an even and b^{3}is an even term.So the result of a

^{3}+ab^{2}+b^{3}gives an odd number but zero is an even number. So there is a contradiction here too.

Case 3:Let’s look into other case where both a, b are odd numbers with no common factors.

Then a

^{3}is an odd term, ab^{2}is an odd and b^{3}is also an odd term.On solving the equation->(1) by substituting the odd terms gives the result an odd number but zero is an even number. So there is a contradiction.

All our assumption leads to contradiction.

So,

there is no rational number r for r^{3}+ r + 1 = 0.