# Numerical analisys close numbers question

nhrock3
$$y=\sqrt{x^2+1}-1$$
how do we know that there is a subtraction of two close numbers thus making loss of significance?
there is no close numbers there is variable X
it could give use a close result or otherwise

and why multiplying and dividing by $$y=\sqrt{x^2+1}+1$$
makes it go away?

## Answers and Replies

Mentor
$$y=\sqrt{x^2+1}-1$$
how do we know that there is a subtraction of two close numbers thus making loss of significance?
there is no close numbers there is variable X
it could give use a close result or otherwise

and why multiplying and dividing by $$y=\sqrt{x^2+1}+1$$
makes it go away?

If x is very close to 0, you're subtracting 1 from a number that is very close to 1.

nhrock3
thanks :)

nhrock3
why multiplying and dividing by $$y=\sqrt{x^2+1}+1$$

Mentor
What do you get when you do this multiplication?
$$\left(\sqrt{x^2+1}-1\right)\frac{\sqrt{x^2+1}+1}{\sqrt{x^2+1}+1}$$

nhrock3
we get x^2 in the nominator
and a sum in the denominator
now there is some stuff about relative error
but i dont know what?