If #n# is a natural number, prove that the number #(sqrt(n) + sqrt(n+1))# is irrational?
2 Answers
See explanation...
Explanation:
It seems that you are not counting
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
#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
#mx+x^2 = m^2+mx-1#
Subtract
#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:
Hence for any positive integer
Hence
Hence
See below.
Explanation:
Supposing
and then
Now solving the system
we get
Concluding if we suppose that