prove that the following are irrational (i) route 4 + route 6
Answers
Step-by-step explanation:
Given: Number √6
To Prove: Root 6 is irrational
Proof: Let us assume that square root 6 is rational. As we know a rational number can be expressed in p/q form, thus, we write, √6 = p/q, where p, q are the integers, and q is not equal to 0. The integers p and q are coprime numbers thus, HCF (p,q) = 1.
√6 = p/q
⇒ p = √6 q ------- (1)
On squaring both sides we get,
⇒ p2 = 6 q2
⇒ p2/6 = q2 ------- (2)
If k was a prime number and k divides a2 evenly, then k also divides 'a' evenly, where a is any positive integer.
Hence the equation (2) shows 6 is a factor of p2 which implies that 6 is a factor of p.
We can write p = 6b (where b is a constant)
Substituting p = 6b in (2), we get
(6b)2/6 = q2
⇒ 36b2/6 = q2
⇒ 6b2 = q2
⇒ b2 = q2/6 ------- (3)
Hence, similarly, equation (3) shows that 6 is a factor of q
We can see from equations (2) and (3) that 6 is a factor of p and 6 is a factor of q respectively which contradicts our assumption that p and q are coprime numbers. Therefore we can conclude that our assumption of taking root 6 as a rational number was wrong.
Thus, the square root of 6 is irrational.
Answer:
2+√6
Step-by-step explanation:
√4+√6
√2^2+√6
(√2)+√6
2+√6