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How do you show that √7 is irrational ?
{Prove}
Time limit : 7 mins
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Answers
Answer:
If possible for a moment let us assume /7 is a rational number..
Hence
/7=a/b
(where a and b are co prime and b =/= 0)
Squaring both sides
7 = a²/b²
7b²=a² - - - - - - - - - - - 1
Hence 7 is the factor of a....
Let us assume..
a=7m (As 7 is a factor of a)
Putting it in equation 1 we get
7b²=49m²
7m²=b²
Hence 7 is also a factor of b...
Hence our assumption is wrong
Hence /7 is an irrational number
Step-by-step explanation:
Co-prime have only one common factor i.e 1...But here 7 is also a common factor...
Hence our assumption is wrong
/7 is irrational...
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Answer:
Step-by-step explanation:
To prove that √7 is irrational
Solution :-
Let us assume that √7 is rational
Then there exists integers a and b such that b not equals to 0 and a,b are co primes
√7 = a/b where b not equals to and (a,b)=1
√7b = a
Squaring on both sides,
(√7b)^2 = a^2
=> 7|a^2 (since ,7|b^2 and a^2 = 7b^2)
=>7|a ——-(1)
So a can be written as
a = 7c for some integer ‘c’
=> a^2 = 7c^2
=>7b^2 = 49 c^2 (since ,a^2 = 7b^2)
=>b^2 = 7c^2
=>7|b^2 [since ,7|7c^2 and b^2 = 7c^2]
=> 7|b ——-(2){since ,by theorem}
From (1) and (2) we have ,
7|a and 7|b .This shows that 7 is a common factor for a and b.
This contradicts the fact that a and b are co-primes
This contraction has arisen because of our wrong assumption that √7 is rational
This our assumption is wrong
Hence √7 is irrational