prove that √3+√5 is an irrational number
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let root 3 + roof 5 be rational
root 3 + root 5 = P/q
(root 3 + root 5) sq=(P/q)sq
3 +5 + 2 root 15 = P sq/q Sq
root I5 = (Psq / qsq -7) 1/2
RHS is rational as all are integers
⇒ LHS is also rational but root 15 is irrational
⇒ root3 + root 5 is irrational.
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Answer:
Step-by-step explanation:
PROVE THAT √3+√5 is an irrational number
Solⁿ :
Let us assume that √3+√5 is a rational number
√3+√5 = a/b (a and b are integers and a and b are co- prime )
a/b - √3 = √5
squaring on both sides
(a/b - √3)² = (√5 )²
⇒ a²/b² - 2√3 a + 3 = 5
b
⇒ a²/b² - 2 = 2√3 a
b
a² - 2b² = 2a √3
b ² b
⇒ a² - 2b² = √3
2ab
⇒ √3 is rational number
But this contradicts the fact that √3 is irrational
So our assumption is incorrect
HENCE WE CONCLUDE THAT √3+√5 IS IRRATIONAL NUMBER.