Question

show that 3+2√7is irrational given that√7is irrationall​

Answer 1

Given:

$\sqrt{7}$ is irrational.

To proof:

$(3+2\sqrt{7})$ as an irrational number

SOLUTION:

Let us assume that $(3+2\sqrt{7})$ as a rational number

So,

$(3+2\sqrt{7}) = \frac{p}{q}$

[In which 'p' and 'q' are integers, (q≠0)]

$\implies 2\sqrt{7}= \frac{p}{q}-3$

[By transporting 3 to RHS]

$\implies 2\sqrt{7}=\frac{p-3q}{q}$

$\implies \sqrt{7} = \frac{p-3q}{q} \div 2$

[By transporting 2 to RHS]

$\implies \sqrt{7}=\frac{p-3q}{q} \times \frac{1}{2}$

$\implies \sqrt{7}=\frac{p-3q}{2q}$

Since, (p), (3), (q), (2) are integers so,

$\boxed{\frac{p-3q}{2q}}$ is a rational number.

But, given that $\sqrt{7}$ is an irrational number and above they are equal.

Thus,

They are not equal.

Our assumption is wrong.

Hence,

$(3+2\sqrt{7})$  as an irrational number.

Answer 2

$\huge \mathbb{ \red {★᭄ꦿ᭄S} \pink{ᴏ}\purple{ʟᴜ} \blue {ᴛ} \orange{ɪ} \green{ᴏɴ★᭄ꦿ᭄}}$

$Given:</p><p>\sqrt{7}7 is \\ irrational.$

$To \: proof:</p><p>(3+2\sqrt{7})as \: an \: irrational \: number</p><p>$

$Let \: us \: assume \: that \: (3+2\sqrt{7})as \: a \: rational \: number...$

$(3+2\sqrt{7}) = \frac{p}{q}(3+27)=qp$

[In which 'p' and 'q' are integers, (q≠0)]

$\implies 2\sqrt{7}= \frac{p}{q}-3$

$\implies 2\sqrt{7}=\frac{p-3q}{q}$

$\implies \sqrt{7} = \frac{p-3q}{q} \div 2$

$\implies \sqrt{7}=\frac{p-3q}{q} \times \frac{1}{2}$

$\implies \sqrt{7}=\frac{p-3q}{2q}$

[By transporting 2 to RHS]

$\boxed{\frac{p-3q}{2q}} is \: a \: rational \: number.$

Since, (p), (3), (q), (2) are integers so,

$But, given that \sqrt{7}7$

is an irrational number and above they are equal.

Thus,

They are not equal.

Our assumption is wrong.

Hence,

(3+27)  as an irrational number.