Question

prove that 3 + √5 is an irrational number​

Answer 1

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Real Numbers

Revisiting Irrational Numbers

prove that 3 + √(5) is an ...

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Asked on December 20, 2019 by

Papita Pushpa

prove that 3+

5

is an irrational number.

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ANSWER

Let us assume that 3+

5

is a rational number.

3+

5

=

b

a

[Here a and b are co-prime numbers]

5

=[(

b

aa

)−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

Answer 2

Answer:

Let us assume that 3+√5 is a rational number equal to 'x'.

$(3+\sqrt{5})^2 or (x)^2=9+5+6\sqrt{5} \\x^2=14+6\sqrt{5} \\This \ implies \ that \ (3+\sqrt{5})^2 \ or \ x^2 \ or \ 14+\sqrt{5}\ is \ a \ rational \ number.\\\sqrt{5}=\frac{x^2-14}{6}\\ This \ implies \ that \ \sqrt{5} \ is \ a \ rational \ number.\\But \ \sqrt{5} \ is\ not \ a \ rational \ number.\\So , \ our\ supposition \ that \ 3+\sqrt{5} \ is \rational\ is \ wrong.\\Hence, \ proved.$

*Square of a rational number is also a rational number.