Math, asked by pradeepkulshrestha19, 1 year ago

Prove that √7 is an irrational number. Hence show that 3+√7 is also an irrational number

Answers

Answered by aakshayagarwal1
5

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

Answered by nitinkumarrrr2p5fpm0
15

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

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