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Real Numbers

    Revisiting Irrational Numbers

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10th class > 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.

Recall: 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, 0.10110111011110 . . . . etc.

To prove that a number like 2 is irrational we need some help. We first need to prove the following:

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 .

Thus, a number a can be expressed as:

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

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

It has been given that: p divides a2. That means from Fundamental Theorem of Arithmetic, p is a of a2 and p is among the factors: p1 p2 p3...pn.

But we assumed a = p1 p2 p3...pn. Since, we know that the prime factors for a number are always :

Prime factors of a2 are p1 p2 p3...pn. Thus, is one of p1 p2 p3...pn.

From this we can clearly see 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.

2=pq

  • Let's assume: 2 is rational. Thus, 2 = (p/q) 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 ,2=ab where a and b are .
  • Squaring on both sides of the equation, we get: 2 = .
  • This implies that since, 2 divides a2 which means divides a as well.
  • If 2 divides a then 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

Let us see for 3.

Instructions

Prove that 3 is irrational

3=pq

  • Let's assume: 3 is rational. Thus, 3 = (p/q) 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 = .
  • 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 tells us further
  • 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 irrational.

Let's prove some of these results.

Instructions

Show that 5−3 is irrational

Prove the above

  • 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, we get
  • 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 [[irrational|rational].

Note:Definition of Coprimes

Moving on to the next question.

Instructions

Show that 32 is irrational

Prove the above

  • 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, we get
  • We can see that a3b is . This also implies that 2 is .
  • But this contradicts the fact that 2 is [[irrational|rational].
  • Thus, 32 is [[irrational|rational].

Note:Definition of Coprimes