Question

prove that root 5 + 2 root 3 is an irrational number​

Answer 1

TO PROVE :- The following as Irrational

$5 + 2 \sqrt{3}$

PROOF

We can easily PROVE it by contradiction method.

If the given number is Rational then ,

Let the number = p/q

where p and q are integers.

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

Here In numerator of RHS one rational number get subtracted from another rational number and hence the result is divided by a rational number therefore, RHS is Rational.

But LHS having root3 is Irrational.

Which contradicts our assumption that P and q are not integers.

Hence, the given number is Irrational.

Answer 2

Step-by-step explanation:

To prove :-

$\sqrt{5} + 2 \sqrt{3} \: is \: irrational$

Proof:-

Let us assume that root 5 + 2 root 3 is a rational number then it can be expressed in the form of a/b, where a,b are co-prime integers and b is not equal to zero.

Now,

$\sqrt{5 } + 2 \sqrt{3} = \frac{a}{b}$

On squaring both side

${( \sqrt{5} + 2 \sqrt{3} )}^{2} = {( \frac{a}{b} )}^{2}$

${( \sqrt{5})}^{2} + {(2 \sqrt{3} )}^{2} + 2( \sqrt{5})(2 \sqrt{3} ) = \frac{ {a}^{2} }{ {b}^{2} }$

$5 + 12 + 4 \sqrt{15} = \frac{ {a}^{2} }{ {b}^{2} }$

$17 + 4 \sqrt{15} = \frac{ {a}^{2} }{ {b}^{2} }$

$4 \sqrt{15} = \frac{ {a}^{2} }{ {b}^{2} } - 17$

$4 \sqrt{15 } = \frac{ {a}^{2} - 17 {b}^{2} }{ {b}^{2} }$

$\sqrt{15} = \frac{ {a}^{2} - 17 {b}^{2} }{4 {b}^{2} } \: \: - - \: - (1)$

by eq-(1) root 15 is rational number.

This contradicts the fact that that root 15 is irrational.

This contradiction is arise due to our wrong assumption that root 5 + 2 root 3 is rational.

Therefore, root 5 + 2 root 3 is irrational.

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