If #n# is a natural number, prove that the number #(sqrt(n) + sqrt(n+1))# is irrational?

2 Answers
Feb 24, 2018

See explanation...

Explanation:

It seems that you are not counting #0# as a natural number, since otherwise it would be a counterexample. Some authors count #0# as a natural number and some do not.

So we assume that "natural number" means positive integer...

Note that:

#(sqrt(n)+sqrt(n+1))^2 = n+2sqrt(n(n+1))+(n+1)#

#color(white)((sqrt(n)+sqrt(n+1))^2) = 2n+1+sqrt(4n^2+4n+1-1)#

#color(white)((sqrt(n)+sqrt(n+1))^2) = 2n+1+sqrt((2n+1)^2-1)#

Suppose #m > 1# is an integer and #x > 0# satisfies:

#x = m-1+1/(1+1/(m-1+x))#

#color(white)(x) = m-1+(m-1+x)/(m+x)#

#color(white)(x) = ((m^2-m+mx-x)+(m-1+x))/(m+x)#

#color(white)(x) = (m^2+mx-1)/(m+x)#

Multiply both ends by #(m+x)# to get:

#mx+x^2 = m^2+mx-1#

Subtract #mx# from both sides to get:

#x^2 = m^2-1#

So:

#x = sqrt(m^2-1)#

So we have found:

#sqrt(m^2-1) = m-1+1/(1+1/(2m-2+1/(1+1/(2m-2+...))))#

In continued fraction notation we can write:

#sqrt(m^2-1) = [m-1;bar(1,2m-2)]#

Since this continued fraction does not terminate, it is not rational.

So: #sqrt(m^2-1)# is irrational.

Hence for any positive integer #n#, #sqrt((2n+1)^2-1)# is irrational.

Hence #(sqrt(n)+sqrt(n+1))^2# is irrational.

Hence #sqrt(n)+sqrt(n+1)# is irrational.

Feb 24, 2018

See below.

Explanation:

Supposing #sqrt(n+1)+sqrtn = p/q# rational fraction, then

#(sqrt(n+1)+sqrtn )(sqrt(n+1)-sqrtn ) = n+1-n = 1 =(sqrt(n+1)-sqrtn ) p/q #

and then

#sqrt(n+1)-sqrtn = q/p# hence #sqrt(n+1)-sqrtn# is rational also.

Now solving the system

#{(sqrt(n+1)+sqrtn = p/q),(sqrt(n+1)-sqrtn = q/p):}#

we get

#sqrt(n+1) = 1/2(p/q+q/p)# is rational and

#sqrtn = 1/2(p/q-q/p)# is also rational

Concluding if we suppose that #sqrt(n+1)+sqrtn# is rational, as a consequence any number #sqrt n# for #n in NN^+# should be also rational which is an absurd.