Question

Show that 1/3 + √7 is an irrational number.

Answer 1

$\sf Let's\ assume\ that\ \dfrac{1}{3}\ +\ \sqrt{7}\ is\ a\ rational\ number.\\\\\\\dfrac{1}{3}\ +\ \sqrt{7}\ =\ \dfrac{a}{b},\ where\ 'a'\ and\ 'b'\ are\ co\ -\ prime\ integers\ and\ b\ is\ \ne\ 0.$

$\sf \dfrac{1}{3}\ +\ \sqrt{7}\ =\ \dfrac{a}{b}\\\\\\\sqrt{7}\ =\ \dfrac{a}{b}\ -\ \dfrac{1}{3}\\\\\\\sqrt{7}\ =\ \dfrac{3a\ -\ b}{3b}\\\\\\Since\ 'a'\ and\ 'b'\ are\ integers\ and\ \dfrac{3a\ -\ b}{3b}\ is\ rational,\\\\\\\implies\ \sqrt{7}\ is\ rational.\\\\\\This\ contradicts\ the\ fact\ that\ \sqrt{7}\ is\ irrational.\\\\\\This\ contradiction\ has\ arisen\ due\ to\ our\ wrong\ assumption.\\\\\\\therefore\ Our\ assumption\ is\ wrong.\\\\\\\dfrac{1}{3}\ +\ \sqrt{7}\ is\ irrational.$

Answer 2

Step-by-step explanation:

Answer

We have to prove that 3+

7

is irrational.

Let us assume the opposite, that 3+

7

is rational.

Hence 3+

7

can be written in the form

b

a

where a and b are co-prime and b



=0

Hence 3+

7

=

b

a

⇒

7

=

b

a

−3

⇒

7

=

b

a−3b

where

7

is irrational and

b

a−3b

is rational.

Since,rational



= irrational.

This is a contradiction.

∴ Our assumption is incorrect.

Hence 3+

7

is irrational.

Hence proved.

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