Prove that 2by√7 is an irrational no.
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assume 2/√7 is rational.
therefore there exist coprime integer a &b such that
2/√7=a/b where b isn't equal to zero.
thus....√7= 2b/a
we know that √7 is irrational
therefore it is not equal to 2b/a
this contradiction because of assumption of 2/√7 as rational ...hence our assumption is wrong
therefore ..2/√7 is irrational
hope it helps!!
therefore there exist coprime integer a &b such that
2/√7=a/b where b isn't equal to zero.
thus....√7= 2b/a
we know that √7 is irrational
therefore it is not equal to 2b/a
this contradiction because of assumption of 2/√7 as rational ...hence our assumption is wrong
therefore ..2/√7 is irrational
hope it helps!!
Answered by
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◙◙ Concept Used : Proof by Contradiction ^_^
♦♦ Prove : : ( 2 / √7 ) is irrational
→ Assume : √7 is rational
=> √7 can be written in the simplest form : ( p / q ) where q ≠ 0
NOTE : → ( p , q ) = 1
=> ( p² / q² ) = 7
=> p² = 7q²
♣ Since, q² is a perfect square, and p² is also a perfect square, 7 has to fit somewhere in the factor of : p²
♥ This can be shown as how we write the factorization of a number ♥
→ Now, 7 divides p² => 7 divides 'p' How ?
♠ Suppose, 7 does not divide p => 7 is not a factor of ( p )² = p²
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Now, 7 divides p
→ Substitute p = 7k in the equation ( because 7 is a factor of p )
=> ( 7k )² = 7q
=> 7k² = q²
=> q² = 7k²
♦ Applying the same logic : 7 divides q² and hence, 7 divides q
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♀ Point to be noted ♀
→ ( p , q ) = 1 | but, both p and q is shown to be divisible by 7
=> Our assumption was Wrong => √7 is not rational
=> √7 is irrational
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√√ We're almost done
→ ( Rational ÷ Irrational is always Irrational ) ←
Proof :
→ Let ( 2 / √7 ) be rational
=> ( 2 / √7 ) = A , where A is a rational number
=> ( 2 / A ) = √7
Checking the Homogeneity of the Equation,
→ ( 2 / A ) on the R.H.S. is Rational [ ∵ Rational ÷ Rational = Rational ]
→ while ( √7 ) as we just proved is Irrational
=> Our Assumption arises a Contradiction
=> ( 2 / √7 ) ≠ Rational
Hence, we proved ( 2 / √7 ) is an Irrational Number.
___________________________________________________________
◙◙ Concept Used : Proof by Contradiction ^_^
♦♦ Prove : : ( 2 / √7 ) is irrational
→ Assume : √7 is rational
=> √7 can be written in the simplest form : ( p / q ) where q ≠ 0
NOTE : → ( p , q ) = 1
=> ( p² / q² ) = 7
=> p² = 7q²
♣ Since, q² is a perfect square, and p² is also a perfect square, 7 has to fit somewhere in the factor of : p²
♥ This can be shown as how we write the factorization of a number ♥
→ Now, 7 divides p² => 7 divides 'p' How ?
♠ Suppose, 7 does not divide p => 7 is not a factor of ( p )² = p²
___________________________________________________________
Now, 7 divides p
→ Substitute p = 7k in the equation ( because 7 is a factor of p )
=> ( 7k )² = 7q
=> 7k² = q²
=> q² = 7k²
♦ Applying the same logic : 7 divides q² and hence, 7 divides q
____________________________________________________________
♀ Point to be noted ♀
→ ( p , q ) = 1 | but, both p and q is shown to be divisible by 7
=> Our assumption was Wrong => √7 is not rational
=> √7 is irrational
___________________________________________________________
√√ We're almost done
→ ( Rational ÷ Irrational is always Irrational ) ←
Proof :
→ Let ( 2 / √7 ) be rational
=> ( 2 / √7 ) = A , where A is a rational number
=> ( 2 / A ) = √7
Checking the Homogeneity of the Equation,
→ ( 2 / A ) on the R.H.S. is Rational [ ∵ Rational ÷ Rational = Rational ]
→ while ( √7 ) as we just proved is Irrational
=> Our Assumption arises a Contradiction
=> ( 2 / √7 ) ≠ Rational
Hence, we proved ( 2 / √7 ) is an Irrational Number.
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