Is #4+sqrt7# rational?

2 Answers
Mar 10, 2018

Answer:

#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)#

Mar 10, 2018

Answer:

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)#