Question

show that 2root 3 -5 is an irrationumber
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Answer 1

Answer:

Step-by-step explanation:

Yes. 2√3-3 is an irrational number

Hope it helps you

Answer 2

Let us assume that 2root3-5 is a rational number.

Hence, it can be expressed in the form of p/q where p and q are integers and q is not equal to 0 and both p and q are co-primes.

Now,

$2 \sqrt{3} - 5 = \frac{p}{q} \\ \\ {(2 \sqrt{3} - 5)}^{2} = \frac{ {p}^{2} }{ {q}^{2} } \\ \\ \\12 + 25 - 20 \sqrt{3} = \frac{ {p}^{2} }{ {q}^{2} } \\ \\ \\20 \sqrt{3} = 37 - \frac{ {p}^{2} }{ {q}^{2} } \\ \\ \\20 \sqrt{3} = \frac{37 {q}^{2} - {p}^{2} }{ {q}^{2} } \\ \\ \\ \sqrt{3} = \frac{37 {q}^{2} - {p}^{2} }{20 {q}^{2} }$

But we know that root 3 is a rational number while the RHS of the equation is a rational number.

This is not possible!

Hence, it is a contradiction which has arose because we took 2root3-5 as rational number.

Hence,

2root3- 5 is an irrational number.

Hence proved!