Question

Is `4+sqrt7` rational?

Answer 1

`sqrt7` is irrational.

Therefore `4+ sqrt7` will also be irrational.

Explanation:

The answer will be irrational.

`sqrt7` is an an irrational number, meaning that it is an infinite non-recurring decimal which cannot be written as a common fraction.

`sqrt7 =2.64575131......`

If you use an irrational number in an operation, the answer will be irrational.

Note that `( sqrt7^2 =7)`

Answer 2

No

Explanation:

If `4+sqrt(7) = p/q` for some integers `p, q` then `sqrt(7) = (p-4q)/q` would also be rational.

To see that `sqrt(7)` is irrational, we can proceed as follows...

Suppose `x > 0` satisfies:

`x = 2+1/(1+1/(1+1/(1+1/(2+x))))`

Then:

`x = 2+1/(1+1/(1+1/(1+1/(2+x))))`

`color(white)(x) = 2+1/(1+1/(1+(2+x)/(3+x)))`

`color(white)(x) = 2+1/(1+(3+x)/(5+2x))`

`color(white)(x) = 2+(5+2x)/(8+3x)`

`color(white)(x) = (21+8x)/(8+3x)`

Multiplying both ends by `(8+3x)` we find:

`3x^2+8x = 21+8x`

Subtracting `8x` from both sides we find:

`3x^2=21`

Hence:

`x^2 = 7`

So:

`x = sqrt(7)`

We have found:

`sqrt(7) = 2+1/(1+1/(1+1/(1+1/(2+sqrt(7)))))`

`color(white)(sqrt(7)) = 2+1/(1+1/(1+1/(1+1/(4+1/(1+1/(1+1/(1+1/(4+...))))))))`

Since this continued fraction does not terminate, it does not represent a rational number.

So `sqrt(7)` is irrational and so is `4+sqrt(7)`