# is $y = \sqrt{x^2 + 1}− x$ a injective (one-to-one) function?

I have a function $y = \sqrt{x^2 + 1}− x$ and I need to prove if it's a Injective function (one-to-one). The function f is injective if and only if for all a and b in A, if f(a) = f(b), then a = b

$\sqrt{a^2 + 1} − a = \sqrt{b^2 + 1} − b$

$\sqrt{a^2 + 1} + b = \sqrt{b^2 + 1} + a$

$(\sqrt{a^2 + 1} + b)^2 = (\sqrt{b^2 + 1} + a)^2$

$a^2 + 1 + b^2 + 2b\sqrt{a^2 + 1} = b^2 + 1 + a^2 + 2a\sqrt{b^2 + 1}$

$2b\sqrt{a^2 + 1} = 2a\sqrt{b^2 + 1}$

$b\sqrt{a^2 + 1} = a\sqrt{b^2 + 1}$

$(b\sqrt{a^2 + 1})^2 = (a\sqrt{b^2 + 1})^2$

$b^2(a^2 + 1) = a^2(b^2 + 1)$

$b^2a^2 + b^2 = a^2b^2 + a^2$

$b^2 = a^2$

$\sqrt{b^2} = \sqrt{a^2}$

$b = \pm a$

then the function is not injective because $b$ in not equal to $a$, but using online solver the function is injective.

• Hint: the derivative is always negative.
– lulu
May 10, 2016 at 13:03
• The extra solution comes from the squaring. May 10, 2016 at 13:03
• @G-man thanks, could you explain better with an example. Thanks in advance May 10, 2016 at 13:04
• Your proof is not correct. You showed that $$f(a)=f(b) \Rightarrow a= \pm b$$ but actually you didn't explicitely give an example for $f(a)=f(-a)$. The error arises when you square everything (third line): this is not a reversible step. May 10, 2016 at 13:05
• Even though it may be reasonable to assume f is defined as $f:\mathbb{R} \rightarrow\mathbb{R}$, it is better to be unambiguous by specifying the domain and codomain. May 10, 2016 at 14:01

When you wrote $$2b\sqrt{a^2 + 1} = 2a\sqrt{b^2 + 1}$$ this gave you a new condition : $a$ and $b$ must have same sign. (The square roots are positive). Hence $a^2=b^2$ implies $a=b$ in this case.
Hint: $$y=\sqrt{x^2+1}-x=\frac1{\sqrt{x^2+1}+x}$$ Then $$f(a)=f(b) \Leftrightarrow \frac1{\sqrt{a^2+1}+a}=\frac1{\sqrt{b^2+1}+b}\Leftrightarrow$$ $$\Leftrightarrow \sqrt{a^2+1}+a=\sqrt{b^2+1}+b \Leftrightarrow a=b$$
$f(t)=\sqrt{t^2+1}+t -$ strictly increasing
• Thanks @Roman83. I am honest to say I did't understand how to simplify $\sqrt{a^2 + 1}$ and $\sqrt{b^2 + 1}$ May 10, 2016 at 13:09