Provethatroot7
irrationalnumber
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Let us assume that √7 is not an irrational
√7 is a rational
p/q where p,q belongs to integers and q ≠ 0
p,q are co-primes
√7 = p/q
squaring on both sides
( √7 )2 = ( p/q )2
7 = p2/q2
p2 = 7q2
p2 is divisible by 7
p is also divisible by 7
let p = 7k
p2 = 7q2
( 7k )2 = 7q2
49k2 = 7q2
7p2 = q2 [ p = 7k ]
q2 = 7p2
q2 is divisible by 7
q is also divisible by 7
"7" is common factor for both p and q
It is contradiction to our assumption
Our assumption is wrong
√7 is an irrational
√7 is a rational
p/q where p,q belongs to integers and q ≠ 0
p,q are co-primes
√7 = p/q
squaring on both sides
( √7 )2 = ( p/q )2
7 = p2/q2
p2 = 7q2
p2 is divisible by 7
p is also divisible by 7
let p = 7k
p2 = 7q2
( 7k )2 = 7q2
49k2 = 7q2
7p2 = q2 [ p = 7k ]
q2 = 7p2
q2 is divisible by 7
q is also divisible by 7
"7" is common factor for both p and q
It is contradiction to our assumption
Our assumption is wrong
√7 is an irrational
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