# Revisiting Irrational Numbers

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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

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- **.

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

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To prove that a number like

**Theorem 1.2:** **Let p be a prime number. If p divides **

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

Thus, a number *a* can be expressed as:

Let **a = **, where

Then we have

It has been **given that:** p divides **from Fundamental Theorem of Arithmetic**, p is a

But we assumed a =

Prime factors of

From this we can clearly see **p divides a.**

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Let us now try to prove **'proof by contradiction'** where we make an assumption and try to prove it.

**Theorem 1.3:**

- Let's assume:
is rational. Thus,2 = (p/q) where p and q are two non-zero integers.2 - Suppose p and q have a common factor other than 1. If we divide by the common factor(s) to get ,
2 = where a and b area b . - Squaring on both sides of the equation, we get: 2 =
. - This implies that since, 2 divides
which meansa 2 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
is divided by 2 which means b is divided byb 2 . - 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,
is2 .

Note:

Let us see for

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**Prove that **

- Let's assume:
is rational. Thus,3 = (p/q) where p and q are two non-zero integers.3 - Suppose p and q have a common factor other than 1. If we divide by the common factor(s) to get ,
3 = where a and b area b . - Squaring on both sides of the equation, we get: 3 =
. - This implies that since, 3 divides
which meansa 2 divides a . - If 3 divides a then we can write a = 3c for some integer c.
- This tells us further
- This means
is divided by 3 which means b is divided byb 2 . - 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,
is3 .

Note:

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.

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**Show that 5 − 3 is irrational**

- Let's assume:
5 − is rational. Thus,3 5 − =3 where p and q are two integers and q ≠ 0. - Dividing by the common factor(s) we get ,
5 − 3 = where a and b area b . - Re-arranging, we get
- We can see that
5 − isa b . This also implies that is3 . - But this contradicts the fact that
is3 . - Thus,
5 − is [[irrational|rational].3

Note:

Moving on to the next question.

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**Show that 3 2 is irrational**

- Let's assume:
3 is rational. Thus,2 3 =2 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 ,
3 2 = where a and b area b . - Re-arranging, we get
- We can see that
isa 3 b . This also implies that is2 . - But this contradicts the fact that
is [[irrational|rational].2 - Thus,
3 is [[irrational|rational].2

Note: