# How to prove an algebraic identity?

I just started a Calc 1 course and I've been asked to prove algebraic identites. I have no clue how to approach this, since I don't even know what an identity is nd how to prove it.

Could someone explain what an algebraic identity is, how to prove an identity, and prove the following identity:

$$\sqrt x - \sqrt y = \dfrac{ (\sqrt x -\sqrt y )(\sqrt x +\sqrt y )}{\sqrt x +\sqrt y }$$

• @herbsteinberg just fixed it, refresh the page Dec 4, 2021 at 18:17
• Term in numerator equals term in denominator - cancel. Net is identity. Dec 4, 2021 at 18:18
• @herbsteinberg so should that be my final answer? My course requires me to articulate every step in detail. Dec 4, 2021 at 18:24
• I can't answer for your course need. Do you need to elaborate on $\frac{A}{A}=1$? Dec 4, 2021 at 18:28
• An algebraic identity is an equality that holds for any values of its variables. This identity only holds when $x,y \neq 0$, otherwise the left-hand side of the equation is $0$ and the right-hand side equals $\frac{0}{0}$. The motivation behind learning this identity in Calculus 1 his is that oftentimes it's easier to take limits when you rationalize the numerator. You prove an identity by demonstrating that one side of the equation always equals the right-hand side by using legitimate algebraic steps. Dec 4, 2021 at 18:36

$$\sqrt{x} - \sqrt{y} = (\sqrt{x} - \sqrt{y})*(\sqrt{x} + \sqrt{y}/(\sqrt{x}+\sqrt{y})$$
$$\Leftrightarrow (\sqrt{x} - \sqrt{y})*(\sqrt{x}+\sqrt{y}) = (\sqrt{x} - \sqrt{y})*(\sqrt{x} + \sqrt{y}$$
$$\Leftrightarrow (\sqrt{x} - \sqrt{y})*(\sqrt{x}+\sqrt{y}) - (\sqrt{x} - \sqrt{y})*(\sqrt{x} + \sqrt{y} = 0$$ which is absolutely true. Thus, the identity holds.