Question

prove √2 is an irrational

Answer 1

$\huge\purple{\mid{\underline{\overline{\texttt{To\:Prove:-}}}\mid}}$

• √2 is an irrational.

$\huge\pink{\mid{\fbox{\tt{PROOF↴}}\mid}}$

→Let us assume on the contrary that

√2 is a rational number.

→Then, there exist positive integers p and q such that √2 = $\frac{p}{q}$

→ [ p and q are in their least terms i.e., HCF of (p,q)=1 and q ≠ 0

On squaring both sides, we get 

                   p²= 2q²      -----(1)                                                                             

Clearly, 2 is a factor of 2q²

⇒ 2 is a factor of p²                                                                    [since, 2q²=p²]

⇒ 2 is a factor of p

 Let p =2 m for all m ( where  m is a positive integer)

Squaring both sides, we get 

            p²= 4 m²    ---------(2)                                                                                     

From (1) and (2), we get 

           2q² = 4m²      ⇒      q²= 2m²

Clearly, 2 is a factor of 2m²

⇒       2 is a factor of q²                                                      [since, q² = 2m²]

⇒       2 is a factor of q 

Thus, we see that both p and q have common factor 2 which is a contradiction that H.C.F. of (p,q)= 1

     Therefore, Our supposition is wrong

Hence √2 is not a rational number i.e., irrational number.

Answer 2

Step-by-step explanation:

Let, if possible, root 2 is rational where p and q are co-prime integers and q is not equal to 0

$\sqrt{2} = \frac{p}{q}$

$= >({ \sqrt{2} })^{2} = \frac{ {p}^{2} }{ {q}^{2} }$

$= > 2 = \frac{ {p}^{2} }{ {q}^{2} }$

$= > 2 {q}^{2} = {p}^{2}$

So, 2q^2 is divisible by 2

So, 2q^2 is divisible by 2i.e. p^2 is divisible by 2

So, 2q^2 is divisible by 2i.e. p^2 is divisible by 2i.e. p is divisible by 2

So, 2q^2 is divisible by 2i.e. p^2 is divisible by 2i.e. p is divisible by 2Let,

$p = 2m$

$= > {p}^{2} = {(2m)}^{2}$

$= > {p}^{2} = 4 {m}^{2}$

$= > 2 {q}^{2} = 4 {m}^{2}$

$= > {q}^{2} = 2 {m}^{2}$

$= > {q}^{2} = 2 {m}^{2}$So, 2m^2is divisible by 2

$= > {q}^{2} = 2 {m}^{2}$So, 2m^2is divisible by 2i.e. q^2 is divisible by 2

$= > {q}^{2} = 2 {m}^{2}$So, 2m^2is divisible by 2i.e. q^2 is divisible by 2i.e. q is divisible by 2

$= > {q}^{2} = 2 {m}^{2}$So, 2m^2is divisible by 2i.e. q^2 is divisible by 2i.e. q is divisible by 2Therefore, p and q have 2 as a common factor other than 1

2 as a common factor other than 1So, it condradicgs our assumption

2 as a common factor other than 1So, it condradicgs our assumptioni.e. our assumption is wrong

2 as a common factor other than 1So, it condradicgs our assumptioni.e. our assumption is wrongSo, root 2 is irrational