Question

prove 3+√5 is irrational
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Answer 1

Answer:

Let us assume that 3+√5 is rational, which means 3+√5 = p/q where p and q are integers and their GCD is 1.

So, squaring both sides we have,

9+5+6√5 = p^2/q^2

So,

6√5 = p^2-14q^2/q^2

Or,

√5 = 1/6(p^2-14q^2/q^2)

Clearly, LHS is irrational since we know that √5 is irrational, this can be clearly proved by elementary means just how we prove that √2 is irrational.

But we see, RHS is a rational number and hence this is not possible and hence a contradiction .

This means our assumption that 3+√5 is rational is incorrect, and hence 3+√5 is irrational.

Answer 2

Answer:

Let us assume ,to contrary ,that 3+√5 is rational.

That is we can find coprime a and b (b is not equal to zero)3+√5

Therefore,3+a upon b = √5

Rearranging this equation,we get √5= 3+a upon b = 3b-a divided by b

since a and b are integers,we get 3+a divided by b is rational ,and so √5 is rational

But this contradict the fact that √5 is rational.

This contradiction has arisen because if our Rong assumption that 3+√5 is rational

so, we conclude that 3+√5 is an irrational number.

Step-by-step explanation:

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