REAL NUMBERS | Irrational Numbers | Part -2

Rational and Irrational Numbers: Concepts and Proofs

Counting Integers Between Two Numbers

To determine how many integers lie strictly between two integers m and n, where m > n, you can use the following formula:

Total count of integers lying strictly between m and n:

m - n - 1

Explanation:

This formula subtracts 1 from the difference between the two numbers because it excludes both endpoints.

Example:

If m = 10 and n = 5:

(10 − 5) − 1 = 4

The integers between 5 and 10 are: 6, 7, 8, and 9 — which gives a total of 4 integers.

Proof That √2 Is Not a Rational Number

We will prove that \( \sqrt{2} \) is not a rational number using a method called proof by contradiction.

Step 1: Assume the opposite

Suppose \( \sqrt{2} \) is a rational number. This means it can be expressed as a ratio of two integers:

\( \sqrt{2} = \frac{a}{b} \)

where \( a \) and \( b \) are integers that have no common factors (i.e., the fraction is in its simplest form), and b ≠ 0 .

Step 2: Square both sides

Squaring both sides of the equation:

\( (\sqrt{2})^2 = \left(\frac{a}{b}\right)^2 \)

\( 2 = \frac{a^2}{b^2} \)

Step 3: Multiply both sides by \( b^2 \)

\( 2b^2 = a^2 \)

This shows that \( a^2 \) is an even number, since it is equal to twice another integer.

Step 4: Deduce that \( a \) is even

If \( a^2 \) is even, then \( a \) must also be even (since the square of an odd number is odd). So we can write:

\( a = 2k \), where \( k \) is some integer.

Step 5: Substitute back into the equation

\( 2b^2 = (2k)^2 = 4k^2 \)

\( b^2 = 2k^2 \)

This means \( b^2 \) is also even, so \( b \) must be even.

Step 6: Contradiction

We have now shown that both \( a \) and \( b \) are even, which means they have a common factor of 2. This goes against our initial assumption that \( a \) and \( b \) share no common divisors.

Assuming that \( \sqrt{2} \) is a rational number results in a logical contradiction. Therefore, \( \sqrt{2} \) is irrational.

Proof That \( \sqrt[3]{6} \) Is Not a Rational Number

We will use proof by contradiction to show that \( \sqrt[3]{6} \) is irrational.

Step 1: Assume the opposite

Suppose \( \sqrt[3]{6} \) is rational. Then it can be written as a ratio of two integers:

\( \sqrt[3]{6} = \frac{a}{b} \), where \( a \) and \( b \) are integers with no common factors and b ≠ 0 .

Step 2: Cube both sides

\( 6 = \left(\frac{a}{b}\right)^3 = \frac{a^3}{b^3} \)

Step 3: Eliminate the denominator by multiplying both sides by \( b^3 \)

\( 6b^3 = a^3 \)

This means that \( a^3 \) is divisible by 6, so \( a \) must also be divisible by both 2 and 3 (i.e., divisible by 6).

Let \( a = 6k \), where \( k \) is an integer.

Step 4: Substitute \( a = 6k \) into the equation

\( 6b^3 = (6k)^3 = 216k^3 \)

\( b^3 = 36k^3 \)

This implies that \( b^3 \) is divisible by 6, so \( b \) must also be divisible by 6.

Step 5: Contradiction

We now find that both \( a \) and \( b \) are divisible by 6, which contradicts the original assumption that \( a \) and \( b \) have no common factors.

This contradiction proves that our assumption was false. Therefore, \( \sqrt[3]{6} \) is not a rational number; it is irrational.