prove that √3+√5 is an irrational number
Answers
Answer:
Given√3 + √5
To prove:√3 + √5 is an irrational number.
Let us assume that√3 + √5 is a rational number.
So it can be written in the form a/b
√3 + √5 = a/b
Here a and b are coprime numbers and b ≠ 0
Solving
√3 + √5 = a/b
On squaring both sides we get,
(√3 + √5)² = (a/b)²
√3² + √5² + 2(√5)(√3) = a²/b²
3 + 5 + 2√15 = a²/b²
8 + 2√15 = a²/b²
2√15 = a²/b² – 8
√15 = (a²- 8b²)/2b
a, b are integers then (a²-8b²)/2b is a rational number.
Then √15 is also a rational number.
But this contradicts the fact that √15 is an irrational number.
Our assumption is incorrect
√3 + √5 is an irrational number.
Hence, proved.
Step-by-step explanation:
please mark me as brainlist
Answer:
Given√3 + √5
To prove:√3 + √5 is an irrational number.
Let us assume that√3 + √5 is a rational number.
So it can be written in the form a/b
√3 + √5 = a/b
Here a and b are coprime numbers and b ≠ 0
Solving
√3 + √5 = a/b
On squaring both sides we get,
(√3 + √5)² = (a/b)²
√3² + √5² + 2(√5)(√3) = a²/b²
3 + 5 + 2√15 = a²/b²
8 + 2√15 = a²/b²
2√15 = a²/b² – 8
√15 = (a²- 8b²)/2b a, b are integers then (a²-8b²)/2b is a rational number.
Then √15 is also a rational number.
But this contradicts the fact that √15 is an irrational number.
Our assumption is incorrect
√3 + √5 is an irrational number.
Hence, proved.
Step-by-step explanation:
Plz mark as brainliest..