# Numerical analisys close numbers question

$$y=\sqrt{x^2+1}-1$$
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?

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Mark44
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.

thanks :)

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

Mark44
Mentor
What do you get when you do this multiplication?
$$\left(\sqrt{x^2+1}-1\right)\frac{\sqrt{x^2+1}+1}{\sqrt{x^2+1}+1}$$

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?