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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:

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

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    $\begingroup$ @herbsteinberg just fixed it, refresh the page $\endgroup$
    – Serket
    Dec 4, 2021 at 18:17
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    $\begingroup$ Term in numerator equals term in denominator - cancel. Net is identity. $\endgroup$ Dec 4, 2021 at 18:18
  • $\begingroup$ @herbsteinberg so should that be my final answer? My course requires me to articulate every step in detail. $\endgroup$
    – Serket
    Dec 4, 2021 at 18:24
  • $\begingroup$ I can't answer for your course need. Do you need to elaborate on $\frac{A}{A}=1$? $\endgroup$ Dec 4, 2021 at 18:28
  • $\begingroup$ 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. $\endgroup$ Dec 4, 2021 at 18:36

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Algebraic identity is essentially a equation involving symbols (say x, y in the one you mentioned), and the equation holds for all the possible values of the symbols as long as they satisfy the specified conditions (in the the identity you mentioned, there are no requirements for x,y, i.e., x and y can take any values and the equation will hold).

A concrete proof of the identity mentioned can be as follows:

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

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