prove that 3√5 is irrational
Answers
Step-by-step explanation:
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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
=[(
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.
OR
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.