Prove that 3+√5 is irrational number
Answers
Answer:
Step-by-step explanation:
✨Here ia the answer ✨____
Let us assume that 3 + √5 is a rational number.
Now,
3 + √5 = (a ÷ b)
[Here a and b are co-prime numbers]
√5 = [(a ÷ b) - 3]
√5 = [(a - 3b) ÷ b]
Here, {(a - 3b) ÷ b} is a rational number.
But we know that √5 is a irrational number.
So, {(a - 3b) ÷ b} is also a irrational number.
So, our assumption is wrong.
3 + √5 is a irrational number.
Hence, proved.
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How √5 is a irrational number.?
→ √5 = a ÷ b [a and b are co-prime numbers]
b√5 = a
Now, squaring on both side we get,
5b² = a² ........(1)
b² = a² ÷ 5
Here 5 divide a²
and 5 divide a also
Now,
a = 5c [Here c is any integer]
Squaring on both side
a² = 25c²
5b² = 25c² [From (1)]
b² = 5c²
c² = b² ÷ 5
Here 5 divide b²
and 5 divide b also
→ a and b both are co-prime numbers and 5 divide both of them.
So, √5 is a irrational number.
Hence, proved
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Answer:
here is your answer
Step-by-step explanation:
Let us assume that 3 + √5 is a rational number.
Now,
3 + √5 = (a ÷ b)
[Here a and b are co-prime numbers]
√5 = [(a ÷ b) - 3]
√5 = [(a - 3b) ÷ b]
Here, {(a - 3b) ÷ b} is a rational number.
But we know that √5 is a irrational number.
So, {(a - 3b) ÷ b} is also a irrational number.
So, our assumption is wrong.
3 + √5 is a irrational number.
Hence, proved.