Proof of Irrationality

The proof of irrationality demonstrates that certain numbers, like the square root of 2, 3, or 5, cannot be expressed as a fraction where both the numerator and denominator are integers. This is a fundamental concept in number theory, challenging the initial intuition that all numbers can be represented as a ratio. Proving irrationality relies on proof by contradiction, assuming the number *is* rational and then deriving a logical inconsistency.

Formulae

The core idea uses the definition of a rational number:

A number $x$ is rational if and only if $x = \frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Furthermore, we can assume that $\frac{p}{q}$ is in its simplest form (i.e., $p$ and $q$ have no common factors other than 1).

We also use the property of divisibility:

If an integer squared is divisible by a prime number, then the integer itself is divisible by that prime number. For example, if $a^2$ is divisible by 2, then $a$ is also divisible by 2.

Examples

Example-1: Proving the irrationality of $\sqrt{2}$

Assume $\sqrt{2}$ is rational. Then, we can write $\sqrt{2} = \frac{p}{q}$, where $p$ and $q$ are integers and the fraction is in its simplest form (no common factors).

Squaring both sides: $2 = \frac{p^2}{q^2}$

Multiplying both sides by $q^2$: $2q^2 = p^2$

This shows that $p^2$ is an even number (divisible by 2). Therefore, $p$ must also be even (because the square of an odd number is odd). We can write $p = 2k$ for some integer $k$.

Substituting $p = 2k$ into $2q^2 = p^2$: $2q^2 = (2k)^2 = 4k^2$

Dividing both sides by 2: $q^2 = 2k^2$

This implies that $q^2$ is also even, and therefore, $q$ must be even.

But, if both $p$ and $q$ are even, they have a common factor of 2, contradicting our initial assumption that $\frac{p}{q}$ is in its simplest form. Therefore, our initial assumption that $\sqrt{2}$ is rational must be false. Hence, $\sqrt{2}$ is irrational.

Example-2: Proving the irrationality of $\sqrt{3}$

Assume $\sqrt{3}$ is rational. Then, we can write $\sqrt{3} = \frac{p}{q}$, where $p$ and $q$ are integers and the fraction is in its simplest form (no common factors).

Squaring both sides: $3 = \frac{p^2}{q^2}$

Multiplying both sides by $q^2$: $3q^2 = p^2$

This shows that $p^2$ is divisible by 3. Therefore, $p$ must also be divisible by 3. We can write $p = 3k$ for some integer $k$.

Substituting $p = 3k$ into $3q^2 = p^2$: $3q^2 = (3k)^2 = 9k^2$

Dividing both sides by 3: $q^2 = 3k^2$

This implies that $q^2$ is also divisible by 3, and therefore, $q$ must be divisible by 3.

But, if both $p$ and $q$ are divisible by 3, they have a common factor of 3, contradicting our initial assumption that $\frac{p}{q}$ is in its simplest form. Therefore, our initial assumption that $\sqrt{3}$ is rational must be false. Hence, $\sqrt{3}$ is irrational.

Theorem with Proof

Theorem: For any prime number $n$, $\sqrt{n}$ is irrational.

Proof:

Assume $\sqrt{n}$ is rational. Then, we can write $\sqrt{n} = \frac{p}{q}$, where $p$ and $q$ are integers and the fraction is in its simplest form (no common factors).

Squaring both sides: $n = \frac{p^2}{q^2}$

Multiplying both sides by $q^2$: $nq^2 = p^2$

This implies that $p^2$ is divisible by $n$. Since $n$ is prime, it follows that $p$ must also be divisible by $n$. We can write $p = nk$ for some integer $k$.

Substituting $p = nk$ into $nq^2 = p^2$: $nq^2 = (nk)^2 = n^2k^2$

Dividing both sides by $n$: $q^2 = nk^2$

This implies that $q^2$ is divisible by $n$, and therefore, $q$ must also be divisible by $n$.

But, if both $p$ and $q$ are divisible by $n$, they have a common factor of $n$, contradicting our initial assumption that $\frac{p}{q}$ is in its simplest form. Therefore, our initial assumption that $\sqrt{n}$ is rational must be false. Hence, $\sqrt{n}$ is irrational.

Common mistakes by students

* Incorrectly Assuming Simplest Form: Not explicitly stating and understanding the implication of the fraction $\frac{p}{q}$ being in its simplest form. This is crucial for the contradiction.

* Misunderstanding Divisibility Rules: Failing to grasp why if $p^2$ is divisible by a prime, then $p$ is also divisible by that prime.

* Logical Errors in the Contradiction: Not clearly explaining how the derived contradiction (e.g., both $p$ and $q$ having a common factor) invalidates the original assumption.

* Generalizing Without Proof: Students may incorrectly assume other roots or combinations are irrational without going through the rigorous proof. For example, assuming $\sqrt{2} + \sqrt{3}$ is irrational without proof.

Real Life Application

While proving irrationality might seem abstract, it’s crucial in various fields:

* Computer Science: Understanding the limitations of representing numbers in computers (e.g., floating-point arithmetic) and the potential for rounding errors.

* Engineering: Designing structures and systems where precise calculations are vital.

* Cryptography: Number theory, including concepts of irrationality, underpins many cryptographic algorithms.

Fun Fact

The discovery of the irrationality of $\sqrt{2}$ by the Pythagoreans in ancient Greece supposedly caused a crisis. They believed that all numbers were rational. The discovery challenged their fundamental beliefs, and according to legend, the Pythagorean who revealed this secret was drowned at sea!

Recommended YouTube Videos for Deeper Understanding

Q.1 Which of the following methods is NOT typically used in the proof of irrationality of a number like $\sqrt{2}$?

Check Solution

Ans: D

While proof by contradiction, the assumption of a rational form, and properties of integers are common, a convergent series is not directly used.

Q.2 Assume $\sqrt{3}$ is rational. Which of the following statements would be a necessary consequence in the proof by contradiction?

Check Solution

Ans: C

If $\sqrt{3} = p/q$, then $3q^2 = p^2$. This implies that 3 divides $p^2$, and therefore 3 divides p. Substituting $p = 3k$, then $3q^2 = 9k^2$, so $q^2 = 3k^2$. This implies 3 divides q. Thus, 3 would divide both the numerator and denominator, contradicting the assumption that the fraction is in the simplest form.

Q.3 If we assume $\sqrt{5}$ is rational, and we can express it as $p/q$ where p and q are coprime integers. Then we can say that 5 divides which term?

Check Solution

Ans: B

If $\sqrt{5} = p/q$, then $5q^2 = p^2$. This shows that 5 divides p^2. Consequently, 5 divides p.

Q.4 Which of the following best describes the fundamental technique used to prove the irrationality of $\sqrt{2}$?

Check Solution

Ans: D

The standard proof of the irrationality of $\sqrt{2}$ uses proof by contradiction.

Q.5 Let’s say we assume $\sqrt{7}$ is rational and express it as a fraction $a/b$ in its simplest form. What conclusion will lead to a contradiction?

Check Solution

Ans: A

Similar to $\sqrt{3}$ and $\sqrt{5}$, if $\sqrt{7} = a/b$, then $7b^2 = a^2$. This indicates 7 must divide $a^2$ and hence $a$. If $a = 7k$, then $7b^2 = 49k^2$, and $b^2 = 7k^2$, implying that 7 divides $b$, contradicting the assumption that $a/b$ is in its simplest form.

Next Topic: Decimal Expansion of Rational Numbers