Question

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

Answer 1

Step-by-step explanation:

Let

$3 + 2 \sqrt{7}$

be a rational number, say r

Therefore,

$3 + 2 \sqrt{7 } = r \\ or \: 2 \sqrt{7} = r - 3 \\ or \: \sqrt{7} = (r - 3) \div 2$

As, r is rational, therefore, r - 3 is also rational and (r -3)/2 is also rational. Hence,

$\sqrt{7}$

is also rational. But these contradicts that

$\sqrt{7}$

is irrational. Therefore, the above number is irrational.

${proved}$

Answer 2

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.