Question

Proof that (5+3√2) is an irrational number

Answer 1

Answer-

Let's assume that 5+3√2 is a rational number.

Therefore,

5+3√2 can be expressed in the form of p/q where p and q are in their lowest form.

Hence, p & q are co-prime integers.

$5 + 3 \sqrt{2} = \frac{p}{q}$

Squaring both LHS and RHS:-

$(5 + 3\sqrt{2} ){}^{2} = \frac{ {p}^{2} }{ {q}^{2} } \\ 25 + 18 + 30 \sqrt{2} = \frac{ {p}^{2} }{ {q}^{2} } \\ 43 + 30\sqrt{2} = \frac{ {p}^{2} }{ {q}^{2} } \\30\sqrt{2} = \frac{ {p}^{2} }{ {q}^{2} } - 43 \\ 30\sqrt{2} = \frac{ {p}^{2} - 43 {q}^{2} }{ {q}^{2} } \\ \sqrt{2} = \frac{ {p}^{2} - 43 {q}^{2} }{ 30{q}^{2} }$

As p and q are integers so p² and q² are also integers. So,$\frac{ {p}^{2} - 43 {q}^{2} }{ 30{q}^{2} }$ is rational.

Hence, RHS is rational.

Therefore, LHS is also rational because LHS = RHS.

So, √2 is rational.

But this contradicts the fact that √2 is irrational.

This contradiction has arisen because our assumption is wrong.

Therefore,

${\bf{5 + 3 \sqrt{2} = Irrational}}$

Answer 2

$\red{\color{white}{\fcolorbox{cyan}{black}{Answer:-}}}$

Let's assume that 5+3√2 is a rational number.

Therefore,

5+3√2 can be expressed in the form of p/q where p and q are in their lowest form.

Hence, p & q are co-prime integers.

$\red{5 + 3 \sqrt{2} = \frac{p}{q}}$

Squaring both LHS and RHS:-

$\red{\begin{lgathered}(5 + 3\sqrt{2} ){}^{2} = \frac{ {p}^{2} }{ {q}^{2} } \\ 25 + 18 + 30 \sqrt{2} = \frac{ {p}^{2} }{ {q}^{2} } \\ 43 + 30\sqrt{2} = \frac{ {p}^{2} }{ {q}^{2} } \\30\sqrt{2} = \frac{ {p}^{2} }{ {q}^{2} } - 43 \\ 30\sqrt{2} = \frac{ {p}^{2} - 43 {q}^{2} }{ {q}^{2} } \\ \sqrt{2} = \frac{ {p}^{2} - 43 {q}^{2} }{ 30{q}^{2} }\end{lgathered}}$

As p and q are integers so p² and q² are rational.

RHS is rational.

LHS is also rational because LHS = RHS.

So, √2 is rational.

But this contradicts the fact that √2 is irrational.

This contradiction has arisen because our assumption is wrong.

${\underline{\underline{\red{5+3\sqrt{2} = IRRATIONAL}}}}$