Prove that √7 is an irrational number. Hence show that 3+√7 is also an irrational number
Answers
Value of √7 will be irrational as it's value will be no terminating and non recurring then add 3 to the value of √7 and the u will get to know that it's value is also irrational due to the same reason mentioned above for √7
Sol:
let us assume that √7 be rational.
then it must in the form of p / q [q ≠ 0] [p and q are co-prime]
√7 = p / q
=> √7 x q = p
squaring on both sides
=> 7q2= p2 ------> (1)
p2 is divisible by 7
p is divisible by 7
p = 7c [c is a positive integer] [squaring on both sides ]
p2 = 49 c2 --------- > (2)
subsitute p2 in equ (1) we get
7q2 = 49 c2
q2 = 7c2
=> q is divisble by 7
thus q and p have a common factor 7.
there is a contradiction
as our assumsion p & q are co prime but it has a common factor.
so that √7 is an irrational.
proving that 3+√7 is an irrational number
Assume rt 7 is rational
rt 7 = a/b
in reduced form
Squarificate it
(rt7)^2 = (a/b)^2
7 = a^2 / b^2
a^2 = 7 b^2
But if a^2 is a multiple of 7 then it is a multiple of 49, and 'a' is a multiple of 7
a = 7c
(7c)^2 = 7 b^2
49c^2 = 7b^2
7c^2 = b^2
Same argument as before with 'a' being a multiple of 7, we see that b is a multiple of 7 now.
But that means a/b = 7c/ 7d = c/d is further reduced, thus contradicting our premise.
Conclusion...
rt7 is irrational.
3 + rt 7 is therefore also irrational