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Valid Squre root in Knewton practice problem

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by bkw » Wed Feb 23, 2011 10:36 am
Hi,

I have experience this in one of Knewton's practice problems which is stated as a valid square root:
sqrt(a^2) = -a, if a<0

I don't understand. If a is negative, say a = -1 then (a)^2 is positive. And the square root per definition has positive result, no?

Grateful for clarification
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by mk101 » Wed Feb 23, 2011 10:44 am
This has more to do with how a square root function is defined...so this result is by definition of a square root function

sqrt(a^2) = mod of a = a for all a>= 0 and -a for all a<0
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by GMATGuruNY » Wed Feb 23, 2011 11:30 am
bkw wrote:Hi,

I have experience this in one of Knewton's practice problems which is stated as a valid square root:
sqrt(a^2) = -a, if a<0

I don't understand. If a is negative, say a = -1 then (a)^2 is positive. And the square root per definition has positive result, no?

Grateful for clarification
Both you and Knewton's statement are correct:
If a = -1, then √a² = √(-1)² = √1 = 1.
-a = -(-1) = 1.
Thus, when a<0, √a² = -a.
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by bkw » Thu Feb 24, 2011 5:11 am
mk101 wrote:This has more to do with how a square root function is defined...so this result is by definition of a square root function

sqrt(a^2) = mod of a = a for all a>= 0 and -a for all a<0
what do you mean by "mod", modulo %?
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by bkw » Thu Feb 24, 2011 5:13 am
GMATGuruNY wrote:
bkw wrote:Hi,

I have experience this in one of Knewton's practice problems which is stated as a valid square root:
sqrt(a^2) = -a, if a<0

I don't understand. If a is negative, say a = -1 then (a)^2 is positive. And the square root per definition has positive result, no?

Grateful for clarification
Both you and Knewton's statement are correct:
If a = -1, then √a² = √(-1)² = √1 = 1.
-a = -(-1) = 1.
Thus, when a<0, √a² = -a.
Gotcha, Thanks!
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by mk101 » Thu Feb 24, 2011 5:41 am
bkw wrote:
mk101 wrote:This has more to do with how a square root function is defined...so this result is by definition of a square root function

sqrt(a^2) = mod of a = a for all a>= 0 and -a for all a<0
what do you mean by "mod", modulo %?
it implies mod function .. and not modulo ... ΙaΙ = a for a >=0 and -a for a<0 ....
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