prove that root 2 + root 5 is an irrational number
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◾ Let √2 + √5 be a rational number.
==> √2 + √5 = a/b
On Squaring both the sides we have,
==> (√2 + √5)² = (a/b)²
==> √2² + 2(√2)(√5) + √5² = (a/b)²............{using the identity (a + b)² = a² + 2ab + b²}
==> 2 + 2√10 + 5 = a²/b²
==> 2√10 + 7 = a²/b²
==> 2√10 = a²/b² - 7
==> 2√10 = (a² - 7b²)/b²
==> √10 = (a² - 7b²)/2b²
==> Here,
(a² - 7b²)/2b² is a rational number.
But, √10 is an irrational number.
Therefore,
Our assumption is proved wrong.
Hence,
√10 is an irrational number.
==> √2 + √5 is also an irrational number.
HOPE IT HELPS YOU.
(^_^)
HERE'S YOUR ANSWER FRIEND,
◾ Let √2 + √5 be a rational number.
==> √2 + √5 = a/b
On Squaring both the sides we have,
==> (√2 + √5)² = (a/b)²
==> √2² + 2(√2)(√5) + √5² = (a/b)²............{using the identity (a + b)² = a² + 2ab + b²}
==> 2 + 2√10 + 5 = a²/b²
==> 2√10 + 7 = a²/b²
==> 2√10 = a²/b² - 7
==> 2√10 = (a² - 7b²)/b²
==> √10 = (a² - 7b²)/2b²
==> Here,
(a² - 7b²)/2b² is a rational number.
But, √10 is an irrational number.
Therefore,
Our assumption is proved wrong.
Hence,
√10 is an irrational number.
==> √2 + √5 is also an irrational number.
HOPE IT HELPS YOU.
(^_^)
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