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Chapter 1: Real Numbers

    Introduction

    The Fundamental Theorem of Arithmetic

    Revisiting Irrational Numbers

    Exercise 1.1

    Exercise 1.2

    Summary

    Enhanced Curriculum Support

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Chapter 1: Real Numbers > Revisiting Irrational Numbers

Revisiting Irrational Numbers

We have talked about irrational numbers and many of their properties in the earlier class. We also studied about their existence and how the rationals and the irrationals together made up the numbers, how to locate irrationals on the number line.

In this section, we will prove that 2 , 3 , 5 and in general: p is irrational, where p is a prime.

One of the theorems that we use in our proof: is the Fundamental Theorem of Arithmetic.

What Makes a Number Irrational?

A number ‘s’ is called if it cannot be written in the form- pq where p and q are and q ≠ 0.

Some examples of irrational numbers, with which you are already familiar, are :

*2, 3, 15,

* π,

*−23,

*The non-terminating, non-repeating decimal 0.10110111011110 . . . . etc.

To prove that a number like 2 is irrational we need some help. We first need to prove the following theorem, whose proof is based on the Fundamental Theorem of Arithmetic.

Theorem 1.2: Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer.

We know from before that every number can be represented by its .

Representation of a:

A number a can be expressed as:

Let a = p1 p2 p3...pn, where p1 p2 etc are primes (not necessarily distinct).

Square of a:

Squaring a gives:

a2 = (p1 p2 p3...pn)(p1 p2 p3...pn) = (p12 p22 p32...pn2)

Given condition:

It is given that p divides a2.

That means from Fundamental Theorem of Arithmetic,the prime factors of a number are unique.Since p divides a2, p is a of a2 .

Uniqueness of Prime Factors:

From the uniqueness of prime factorization, the prime factors of a2are exactly p1 p2 p3...pn, but each raised to the power 2:

a2 = (p12 p22 p32...pn2)

Thus, is one of p1 p2 p3...pn, which are the prime factors of a.

Conclusion:

Since p is a prime factor of a, p divides a.

Instructions

Let us now try to prove 2 is irrational. We do using 'proof by contradiction' where we make an assumption and try to prove it.

Theorem 1.3: 2 is irrational.

AssumptionContradictionMethod:2=pq

  • Let's assume: 2 is rational. Thus, 2 = pq where p and q are two integers.
  • Suppose p and q have a common factor other than 1. If we divide by the common factor(s) to get ,2=ab where a and b are .
  • Squaring on both sides of the equation, we get: 2 = .
  • Taking b2 on to the LHS.
  • Implication: Since 2b2=a2, a2 is even as it is two times another integer. If a2 is even, a must also be even(because the square of an odd number is odd).
  • This implies that 2 divides a2 which means divides a as well.If 2 divides a then expressing a as an even number, we can write a = 2c for some integer c.
  • Squaring both sides, this further gives us
  • This means b2 is divided by 2 which means b is divided by .
  • Thus, a is divided by 2 and b is divided by 2. But this is not possible as a and b are .
  • This means that our assumption is .
  • Thus, 2 is .

Note: Definition of Coprimes

Instructions

Prove that 3 is irrational

AssumptionContradictionMethod:3=pq

  • Let's assume: 3 is rational. Thus, 3 = pq where p and q are two non-zero integers.
  • Suppose p and q have a common factor other than 1. If we divide by the common factor(s) to get ,3=ab where a and b are .
  • Squaring on both sides of the equation, we get: 3 = .
  • Taking b2 on to the LHS.
  • Implication: Since 3b2=a2, a2 is divisible by 3. From Theorem 1.2, if a prime p divides a2, then p must divide a. Thus, 3 divides a.
  • This implies that since, 3 divides a2 which means divides a.If 3 divides a then we can write a = 3c for some integer c.
  • This means b2 is divided by 3 which means b is divided by .
  • Thus, a is divided by 3 and b is divided by 3. But this is not possible as a and b are .
  • This means that our assumption is .
  • Thus, 3 is .

Note: Definition of Coprimes

In earlier classes, we have learnt that :

The sum or difference of a rational and an irrational number is irrational.

The product and quotient of a non-zero rational and irrational number is .

Let's prove some of these results.

Instructions

Show that 5−3 is irrational

Assumption (Contradiction Method)=5 – sqrt(3)=p/q

  • Let's assume: 5−3 is rational. Thus, 5−3 = where p and q are two integers and q ≠ 0.
  • Dividing by the common factor(s) we get ,5−3=ab where a and b are .
  • Re-arranging to Isolate 3, we get
  • 5b−ab=3
  • We can see that 5−ab is . This also implies that 3 is .
  • But this contradicts the fact that 3 is .
  • Thus, 5−3 is .

Note:Definition of Coprimes

Instructions

Show that 32 is irrational

Assumption (Contradiction Method)-3sqrt(2) is rational

  • Let's assume: 32 is rational. Thus, 32 = where p and q are two integers and q ≠ 0.
  • Suppose p and q have a common factor other than 1. If we divide by the common factor(s) to get ,32=ab where a and b are .
  • Re-arranging to isolate 2, we get
  • Implication:We can see that a3b is . This also implies that 2 is .
  • But this contradicts the fact that 2 is .
  • Thus, 32 is .

Note: Definition of Coprimes