Question

answer this question​

Answer 1

CORRECT QUESTION:

➧ Show that 1/√3 is Irrational.

➧ Prove that √6 is Irrational.

SOLUTION:

❶ Let assume to the contrary, that 1/√3 is rational. That is, we can find co-prime integers p and q (q ≠ 0) such that

$\rm{\rightharpoonup \dfrac{1}{\sqrt{3}} = \dfrac{p}{q}}$

$\rm{\rightharpoonup \dfrac{1 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \dfrac{p}{q}}$

$\rm{\rightharpoonup \dfrac{\sqrt{3}}{3} = \dfrac{p}{q}}$

$\rm{\rightharpoonup \sqrt{3} = \dfrac{3p}{q}}$

Since, p and q are integers so 3p/q is rational, and so √3 is rational.

But this contradicts the fact that √3 is irrational.

So, we conclude that √3 is an irrational.

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SOLUTION:

❷ We assume that √6 is rational number. Then √6 can be expressed in the form of p/q, where p and q are coprimes

$\rm{\rightharpoonup \sqrt{6} = \dfrac{p}{q}}$

Squaring on both sides

$\rm{\rightharpoonup 6= \dfrac{{p}^{2}}{{q}^{2}}}$

$\rm{\rightharpoonup 6q^2 = p^2...1)}$

$\rm{\rightharpoonup 6 \: divides \: p^2}$

$\rm{\rightharpoonup 6 \: divides \: p...2)}$

Let

$\rm{\rightharpoonup p = 6m}$

$\rm{\rightharpoonup p^2 = 36m^2}$

Putting the value of p² in 1), we get

$\rm{\rightharpoonup 6q^2 = 36m^2}$

$\rm{\rightharpoonup q^2 = 6m^2}$

$\rm{\rightharpoonup 6 \: divides \: q^2}$

$\rm{\rightharpoonup 6 \: divides \: q...3)}$

Thus, from (2) p is a multiple of 6 and from (3), q is also a multiple of 6. it means 6 is a common factor of p and q.

Hence, √6 is an irrational number.

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