prove that 2+ root 5is irrational
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Let us assume that √5 is rational.
Where √5 = p / q, p, and q are co-prime and q ≠ 0.
=> √5 = p / q
=> √5 q = p
=> 5 q² = p²
Therefore, p is the factor of 5. ---------------( 1 )
p = 5k ( k = any rational number.)
=> p² = 25 k²
=> 5 q² = 25 k²
=> q² = 5 k²
Therefore, q is the factor of 5. -----------------( 2 )
From( 1 ) and ( 2 )-
p and q are the factors of 5.
which contradicts our assumption that p and q are co-prime.
Therefore, what we assumed was wrong.
Hence, √5 is an irrational number.
Let us assume that 2 + √5 is a rational number.
where it is equal to a / b, here a and b are co-prime and b ≠ 0.
2 + √5 = a / b
=> √5 = a / b - 2
=> √5 = (a - 2b) / b
(a - 2b) / b is a rational number.
which contradicts the fact that √5 is irrational.
Therefore, what we assumed was wrong.
2 + √5 is an irrational number.
Hence proved.
Hope this helps.
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Where √5 = p / q, p, and q are co-prime and q ≠ 0.
=> √5 = p / q
=> √5 q = p
=> 5 q² = p²
Therefore, p is the factor of 5. ---------------( 1 )
p = 5k ( k = any rational number.)
=> p² = 25 k²
=> 5 q² = 25 k²
=> q² = 5 k²
Therefore, q is the factor of 5. -----------------( 2 )
From( 1 ) and ( 2 )-
p and q are the factors of 5.
which contradicts our assumption that p and q are co-prime.
Therefore, what we assumed was wrong.
Hence, √5 is an irrational number.
Let us assume that 2 + √5 is a rational number.
where it is equal to a / b, here a and b are co-prime and b ≠ 0.
2 + √5 = a / b
=> √5 = a / b - 2
=> √5 = (a - 2b) / b
(a - 2b) / b is a rational number.
which contradicts the fact that √5 is irrational.
Therefore, what we assumed was wrong.
2 + √5 is an irrational number.
Hence proved.
Hope this helps.
Please mark the brainliest.
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