Real Number-Revisiting Irrational Numbers


We have studied in the previous class about rational numbers as well as irrational numbers. Both are the members of the REAL NUMBERS family. We have also studied how to locate irrational numbers on the number line.

Any rational number is represented by \displaystyle \frac{p}{q}; where \displaystyle q\ne 0 whereas irrational number is represented by ‘s’

0.10110111011110….. etc. are the examples of irrational numbers.

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

Proof :

Let the prime factorization of a be as follows :

a = p1, p2, p3, p4, p5, …., …. pn

\displaystyle \therefore  a2 = (p1, p2, p3, p4, …. Pn) (p1, p2, p3, p4, …. Pn)

= p12, p22, p32, p42, …………… pn2.

It is given that p divides a2, so from the Fundamental Theorem of Arithmetic p is one of the prime factors of a2. Also a2 having prime factors p1, p2, p3, ….. pn.

Hence, p is one of p1, p2, p3, …… pn.

Example 6: To prove \displaystyle \sqrt{2} is irrational.

Solution. Let us assume that \displaystyle \sqrt{2} is rational.

Now, we can find integers r and \displaystyle s\left( \ne 0 \right) such that \displaystyle \sqrt{2}=\frac{r}{s}.

Again let r and s have a common factor other than 1. Then we can divide by common factor to get, where a and b are coprime.

\displaystyle \Rightarrow   \displaystyle b\sqrt{2}=a …….. (i)

\displaystyle \Rightarrow   \displaystyle {{\left( b\sqrt{2} \right)}^{2}}={{\left( a \right)}^{2}} [squaring both sides]

\displaystyle \Rightarrow  2b2 = a2

\displaystyle \therefore 2 divides a2.

Now, by above theorem

2 divides a also.

So, we can write a = 2c for some integer c.

\displaystyle \Rightarrow   \displaystyle b\sqrt{2}=2c [From (i)]

\displaystyle \Rightarrow  2b2 = 4c2 [Squaring both sides]

\displaystyle \therefore  b2 = 2c2 ……. (ii)

i.e., 2 divides b2

So, 2 divides b [by using above theorem with p = 2]

Again, a and b have at least 2 as a common factor.

But this contradicts the fact that a and b have no common factors other than 1.

This contradiction has arisen because of our incorrect assumption that \displaystyle \sqrt{2} is rational.

Hence, we conclude that \displaystyle \sqrt{2} is irrational.

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