Prove using only integers and inequalities that there is no natural number whose square is 10

Question

(Producing a decimal approximation of √10, or arguing that √10 is between 2 and 3 is not proof. Just because one solution of

Answer 1

If x=√10 and x is not a natural number, then it lies between two consecutive natural numbers: n<x<n+1, where n is a natural number.

So x^2>n^2 and x^2<n^2+2n+1. Therefore, putting x^2=10, we have 10>n^2, i.e., n^2<10, and n^2+2n+1>10, n^2+2n>9.

n(n+2)>9 is satisfied for n≥3. But n^2<10 is only satisfied for n≤3 so n=3 satisfies both conditions. Therefore 3<x<4 and x lies between two consecutive natural numbers.