Time dilation by special relativity When reading about special relativity and time dilation I encounter a problem; Here is a link: Time dilation in GPS On page 1 under header "2. Time dilation by special relativity." It says: Since $(1 – x)^{-1/2} ≈ 1 + x /2$ for small $x$, we get... How is $(1 – x)^{-1/2} ≈ 1 + x /2$ for small $x$? I really don't get it. Thank you in advance!

Taylor approximation

$$f(x)= f(0)+f'(0)\cdot x+\cdots,$$

or, for $f(x)=(1+x)^\alpha$, we have

$$(1+x)^\alpha\approx 1+\alpha x.$$