Prove that 3+ √5 is an irrational number
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Answered by
0
Answer:
Let us assume that 3+
5
is a rational number.
Now,
3+
5
=
b
a
[Here a and b are co-prime numbers]
5
=[(
b
a
)−3]
5
=[(
b
a−3b
)]
Here, [(
b
a−3b
)] is a rational number.
But we know that
5
is an irrational number.
So, [(
b
a−3b
)] is also a irrational number.
So, our assumption is wrong.
3+
5
is an irrational number.
Hence, proved.
I HOPE IT HELP YOU...
Answered by
0
3 + √5 is a irrational number. Hence, proved. ... → a and b both are co-prime numbers and 5 divide both of them. So, √5 is a irrational number.
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