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# Where Is Mistake?

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sqrt(-4) = (-4)^(1/2) = ((-4)^(2))^(1/4) = sqrt((-4)^(2),4) = sqrt(16,4) = 2. Where is mistake?

Apr 30, 2014

#2
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You have made use of the relation xa*b = (xa)b.  But this has a limited range of applicability. In particular it doesn't apply when x is negative and b is a fraction.  If you insist on applying it outside of its range of applicability you will get inconsistent mathematics (as your example illustrates!).

Apr 30, 2014

#1
0

sqrt(-4) = (-4)^(1/2) = ((-4)^(2))^(1/4) = sqrt((-4)^(2),4) = sqrt(16,4) = 2. Where is mistake?

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Note that √(-4) = √(4)*√(-1) =2√(-1)

And by definition,  √(-1) = √(i2) = i

So, √(-4) = 2i  ≠ 2

Apr 30, 2014
#2
+10

You have made use of the relation xa*b = (xa)b.  But this has a limited range of applicability. In particular it doesn't apply when x is negative and b is a fraction.  If you insist on applying it outside of its range of applicability you will get inconsistent mathematics (as your example illustrates!).

Alan Apr 30, 2014
#3
+5

The mistake lies partly in writing 1/2 as 2/4. Squaring an expression often results in the creation of spurious answers, as for example, x=2, x^2=4, x=sqrt(4), x=2 or x=-2.

As to the actual question, to discuss it you have to involve complex numbers.

$$(-4)^{1/2}=(4\angle(180+360k))^{1/2}=2\angle(90+180k),\quad k=0,1.$$

That gets you 2 values for the square root of -4,

$$2\angle 90=2i \text{ or } 2\angle 270=-2i.$$

Now look at the final result as given, sqrt(16,4), which I take to mean the fourth root of 16.

$$16^{1/4}=(16\angle(0+360k))^{1/4}=2\angle 90k, \quad k=0,1,2,3.$$

$$2\angle 0=2,\quad 2\angle90=2i,\quad 2\angle 180=-2 \quad \text{ and }\quad 2\angle 270=-2i,$$