Let x=√2, so x^2=2.
Suppose we don't know how to evaluate √2 or the square root of any number that isn't a perfect square.
But we can observe that the fraction (100-2)/49=2, and 100/49=(10/7)^2. Therefore we can write 2=(10/7)^2(1-0.02). That is, x^2=(10/7)^2(1-0.02) and taking square roots of each side we get x=(10/7)√(1-0.02). Although we can't evaluate the square root, we know it must be a little smaller than 1.
We also observe that (10000+82)/5041=2, and 10000/5041=(100/71)^2. Therefore we can write 2=(100/71)^2(1+0.0082). That is, x^2=(100/71)^2(1+0.0082) and taking square roots of each side we get x=(100/71)√(1+0.0082). Although we can't evaluate the square root, we know it must be a little bigger than 1.
Therefore x<10/7 and x>100/71, or 100/71<x<10/7. Since x=√2 we have shown that the value of √2 lies between 100/71 and 10/7. Thus an irrational number is trapped between two rational fractions---this is a conjecture based on the example: what applies to one irrational number can be conjectured to apply to all such numbers.
Another example is √3 which lies between 12/7 and 7/4; or 20/9<√5<9/4. In general:
na/b<√n<b/a where b/a is an approximation to √n which is greater than √n, i.e., (b/a)^2>n. For example, 14/10<√2<10/7, which is the same as 7/5<√2<10/7, because 49/25<2<100/49. This strengthens the conjecture, because fractions can always be found, however crude the approximation. By taking the average of the limits of the range we arrive at another approximation and we can use this to define another pair of limits, so narrowing the interval for √n. So the average is (na^2+b^2)/(2ab) which will define one of the limits, and hence determine the other. In the case of n=2 we have the next limits as 140/99<√2<99/70. Therefore we have a quadratic formed from x=(na^2+b^2)/2ab or na^2+b^2-2abx=0 which is solved for x rather than a or b, and this gives us progressively accurate values for x=√n as a and b are progressively redefined.