Question

prove that root 3 is irrational​

Answer 1

Answer:

Let us assume on the contrary that 3 is a rational number. Then, there exist positive integers a and b such that

3=ba where, a and b, are co-prime i.e. their HCF is 1

Now,

3=ba

⇒3=b2a2 

⇒3b2=a2 

⇒3∣a2[∵3∣3b2] 

⇒3∣a...(i) 

⇒a=3c for some integer c

⇒a2=9c2 

⇒3b2=9c2[∵a2=3b2] 

⇒b2=3c2 

⇒3∣b2[∵3∣3c2] 

⇒3∣b...(ii)

From (i) and (ii) root 3 is irrational

Explanation:

Let us assume on the contrary that 3 is a rational number. Then, there exist positive integers a and b such that

3=ba where, a and b, are co-prime i.e. their HCF is 1

Now,

3=ba

⇒3=b2a2 

⇒3b2=a2 

⇒3∣a2[∵3∣3b2] 

⇒3∣a...(i) 

⇒a=3c for some integer c

⇒a2=9c2 

⇒3b2=9c2[∵a2=3b2] 

⇒b2=3c2 

⇒3∣b2[∵3∣3c2] 

⇒3∣b...(ii)

From (i) and 

Answer 2

Let us assume root 3 is rational no.
As we know root 3 = p/q where q is not equal to rational no
Root3 =p/q
Squaring on both sides
Cross multiply
P=9qeq1)
Let us assume 3q=c
Where c is quotient
Division