Question

show that
$5 + 2 \sqrt{7}$
is irrational where
$\sqrt{7}$
is given to be an irrational
​

Answer 1

Given :-

$\sqrt{7 \:} \: is \: irrational \:$

Solution :-

$let \: assume \: 5 \: + 2 \sqrt{7 \:} \: is \: \:an \: \\ rational \: number \: \\ therefore \: 5 \: + \: 2 \sqrt{7 } \: = \: a \div b \\ (a \: and \: b \: are \: integers \: )\: \\ 2 \sqrt{7 } \: = \: a \: \div \: b \: - 5 \\ 2 \sqrt{7 } = \: a \: - \: 5b \div b \\ \sqrt{7} \: = \: a \: - \: 5b \: \div 2b \: \\ \: since \: here \: a \: and \: b \: are \: integers \: \: \\ therefore \: \: \sqrt{7} \: is \: rational \: . \\ this \: contradicts \: the \: fact \: that \: \: \\ \sqrt{7} \: is \: irrational \: .therefore \: our \: \\ assumption \: was \: wrong \: \\ hence \: \: 5 \: + \: 2 \sqrt{7} \: is \: irrational \:$

Answer 2

Step-by-step explanation:

Given :-

\sqrt{7 \:} \: is \: irrational \:

7

isirrational

Solution :-

\begin{gathered}let \: assume \: 5 \: + 2 \sqrt{7 \:} \: is \: \:an \: \\ rational \: number \: \\ therefore \: 5 \: + \: 2 \sqrt{7 } \: = \: a \div b \\ (a \: and \: b \: are \: integers \: )\: \\ 2 \sqrt{7 } \: = \: a \: \div \: b \: - 5 \\ 2 \sqrt{7 } = \: a \: - \: 5b \div b \\ \sqrt{7} \: = \: a \: - \: 5b \: \div 2b \: \\ \: since \: here \: a \: and \: b \: are \: integers \: \: \\ therefore \: \: \sqrt{7} \: is \: rational \: . \\ this \: contradicts \: the \: fact \: that \: \: \\ \sqrt{7} \: is \: irrational \: .therefore \: our \: \\ assumption \: was \: wrong \: \\ hence \: \: 5 \: + \: 2 \sqrt{7} \: is \: irrational \: \end{gathered}

letassume5+2

7

isan

rationalnumber

therefore5+2

7

=a÷b

(aandbareintegers)

2

7

=a÷b−5

2

7

=a−5b÷b

7

=a−5b÷2b

sincehereaandbareintegers

therefore

7

isrational.

thiscontradictsthefactthat

7

isirrational.thereforeour

assumptionwaswrong

hence5+2

7

isirrational