Prove that 3+√5 is an irrational number

1.	Prove that 3+√5 is an irrational number
Answer» If POSSIBLE, then let 3+√5 3+√5 = p/q Sqaring both sides, (3+√5)^2 = p^2/q^2 9+5+23√5 = p^2/q^2 14+6√5 = p^2/q^2 14 is a RATIONAL number, 6√5 is an irrational number, and p^2/q^2 is a rational number Then, 6√5 = p^2/q^2 - 14 rational - rational = rational Irrational number is not equal to rational number Then, our ASSUMPTION that 3+√5 is a rational number is wrong Hence, 3+√5 is an irrational number.

Answer»

If POSSIBLE, then let 3+√5

3+√5 = p/q

Sqaring both sides,

(3+√5)^2 = p^2/q^2

9+5+2*3*√5 = p^2/q^2

14+6√5 = p^2/q^2

14 is a RATIONAL number, 6√5 is an irrational number, and p^2/q^2 is a rational number

Then,

6√5 = p^2/q^2 - 14

*rational - rational = rational*

Irrational number is not equal to rational number

Then, our ASSUMPTION that 3+√5 is a rational number is wrong

Hence, 3+√5 is an irrational number.

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