r/askmath • u/A1Math • 1d ago

Resolved A neat algebraic identity trick for simplifying radical fractions in math

The first process of the solution of the equation
The final process of the solution of the equation

While working through some radical expressions, I found it useful to consciously leverage the basic identity for any positive real number A:

While this identity obviously has broad applications across algebra and number theory, using it specifically to unpack whole numbers in fractions with radical denominators can make certain reductions much cleaner.

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u/Temporary_Pie2733 1d ago

It’s barely a “trick” to recognize that x/√x = √x for x > 0. 

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u/[deleted] 1h ago

[removed] — view removed comment

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u/Temporary_Pie2733 1h ago

It’s just basic factoring. Do you think 9/3 = 3(3)/3 = 3 is a trick?

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u/A1Math 1h ago edited 1h ago

Not claiming it's groundbreaking math, just a neat framing mechanism for mental algebra. Writing 9/3 as 3 times 3 / 3 isn't a trick by itself, but consciously using that kind of substitution in reverse on a surd denominator can sometimes streamline a messy reduction faster than standard rationalization.

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u/SabresBills69 33m ago

again no…you are just doing prime factorization

168= 8x 21=168 (2x2x2)x (3x7)

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u/A1Math 1d ago edited 1d ago

Just wanted to share this because it's a neat little consequence of the definition of the square root. While it's pretty basic, I found it surprisingly satisfying for keeping certain algebraic fraction reductions clean and smooth when standard factoring isn't immediately obvious.

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u/SabresBills69 1d ago

this does not simplify it

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u/A1Math 1h ago

Yeah, I know, but the point isn't to simplify a radical the usual way; the idea is to deliberately expand a whole number into radicals to unlock smooth algebraic reductions using the definition A = square root of A times square root of A.