__Revisiting Irrational Numbers__

**Revisiting Irrational Numbers**

**Irrational numbers are those numbers which cannot be written in the form ** **, where p and q are integers and q ≠ 0. E.g.,, **

**The square roots of all the numbers do not give an irrational number.**

**For example, **** **** is an irrational number but = 2, which is rational.**

**Therefore square roots of all prime numbers are irrational.**

**If p is a prime number then is an irrational number.**

**Theorem 1:**

**If a prime number p divides a ^{2}, then p divides a, where a is a positive integer.**

**Proof:**

**Every positive integer can be expressed as the product of primes.**

**Let a = p _{1}p_{2}p_{3} ... ... ... . . p_{n} where p_{1}p_{2}p_{3} ... ... ... . . p_{n} are all the prime numbers of a.**

** a ^{2} = (p1p2p3 ... ... ... . . pn)**

**a ^{2} = (p_{1}p_{2}p_{3} ... ... ... . . pn)(p_{1}p_{2}p_{3 }... ... ... . . p_{n})**

** a ^{2} = (p_{1}^{2}p_{2}^{2}p_{3}^{2} ... ... ... . . pn^{2} )**

**It is given that p divides a ^{2 }. According to the Fundamental Theorem of Arithmetic, we can say that p is one of the prime factors of a^{2}.**

**According to the Fundamental Theorem of Arithmetic, the prime factorization of a natural number is unique.**

**Now the only prime factors of a ^{2 }are p_{1}p_{2}p_{3} ... ... ... . . p_{n}. Therefore, p is one of p^{1}p^{2}p^{3} ... ... ... . . p^{n}.**

**So p is also a factor of a**

**If p divides a ^{2} then p also divides a.**

**Let us consider a positive integer 12.**

**Now, the factors of 12 are 2, 2 and 3.**

**On squaring 12 we get,**

**12 ^{2} = 144**

**The factors of 144 are 2, 2, 2, 2, 3 and 3.**

**If one of the factors, let’s say 3 divides 144. Then this factor, 3 will also divide 144, as it is one of the factors of 144 also.**

**Example: Prove that is an irrational number.**

**Let us assume that **** ****is rational. Since it is a rational number it can be expressed in the form ** **, where a and b are integers and b ≠ 0.**

**Now, a and b have no common factor other than 1.**

** = ** **, where a and b are coprime**

**On squaring both sides, we get
2 = **

** 2b ^{2} = a^{2}**

**Therefore, 2 divides a ^{2}.**

**We know that when 2 divides a ^{2 }then 2 divides a also.**

**We can write a = 2m, where m is an integer.**

**Putting a = 2m in 2b ^{2} = a^{2}**

** 2b ^{2} = (2m)^{2}**

** 2b ^{2} = 4m^{2}**

** b**^{2} = 2m^{2}

**Again if 2 divides b ^{2 }, then 2 divides b also.**

**Therefore, 2 is a common factor of a and b.**

**But this contradicts the fact that a, b have no common factor other than 1.**

**So, we conclude that **** **** is irrational.**

**Example: Show that 3√2 is an irrational number.**

**Let us assume that 3√2 is rational.**

**Then 3√2 can be expressed in the form ** **where a and b are integers and ****b ≠ 0. Now, a and b have no common factor other than 1.**

**3**** ****= ** **, where a and b are coprime integers**

** = **

**Since, 3, a and b are integers, **** ****is rational.**

**Now, a rational number cannot be equal to an irrational number, that is**

**So, we conclude that 3**** is irrational.**

**Example: Show that 5 − is irrational.**

**Let us assume that 5 − is rational. As it is a rational number, it can be**

**expressed in the form ** **where a and b are integers and b ≠ 0. Now, a and ****b have no common factor other than 1.**

**5 − **** =ab, where a and b are coprime**

**5 −****= **

**= **

**As, 5, a, and b are integers, **** is rational.**

**Now, **** is an irrational number and cannot be equal to a rational number.**

**So, we conclude that 5 − **** is irrational.**

**Example: Prove that 2√3 + √5 is an irrational number.**

**Let us assume that 2**** + √5 is rational. As it is a rational number, it can be expressed in the form ** **where a and b are integers and b ≠ 0. Now, a ****and b have no common ****factor other than 1.**

**2**** + **** = ** **, where a and b are coprime**

**2****= ** **− **

**On squaring both sides, we get
12 = ( **

**− √5 )**

^{2}**12 =a ^{2}/b^{2} − 2 **

**+ 5**

** − 2**** ****= 7**

** ****− 7 = 2**** **

** = **** **

** =**

**Since, 2, 7, a and b are integers **** is rational**

**A rational number cannot be equal to an irrational number.**

**So, our assumption is wrong. Hence, 2**** + **** is irrational.**