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 2019-09-27, 21:20 #1 wildrabbitt   Jul 2014 3·149 Posts Can some explain this? I don't understand why $$\sqrt{ab}=\sqrt{a}\sqrt{b}$$ fails for negative a and b. e.g $$\sqrt{(-4)(-9)}=\sqrt{36}=6\neq\sqrt{(-4)(-9)}=\sqrt{(-1)(4)(-1)(-9)}=\sqrt{-1}\sqrt{-1}\sqrt{4*9}=-1\times6$$ Any replies helping me understand this? Last fiddled with by wildrabbitt on 2019-09-27 at 21:21
 2019-09-27, 21:32 #2 VBCurtis     "Curtis" Feb 2005 Riverside, CA 12AB16 Posts The square root of x^2 is |x|, not x. So the square root of (-1)^2 is |-1|, not -1. It's not OK to write sqrt(x^2) = sqrt(x) * sqrt (x) = x, but that's what you did for -1.
 2019-09-27, 21:48 #3 wildrabbitt   Jul 2014 3·149 Posts Thanks very much. I feel relieved.
2019-09-28, 01:50   #4
LaurV
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Quote:
 Originally Posted by wildrabbitt I don't understand why $$\sqrt{ab}=\sqrt{a}\sqrt{b}$$ fails for negative a and b. e.g $$\sqrt{(-4)(-9)}=\sqrt{36}=6\neq\sqrt{(-4)(-9)}=\sqrt{(-1)(4)(-1)(-9)}=\sqrt{-1}\sqrt{-1}\sqrt{4*9}=-1\times6$$ Any replies helping me understand this?
It does not fail, you just wrote it in a not-very-correct way.

$$\sqrt{(-4)\times(-9)}=\sqrt{36}=\pm 6=\sqrt{(-4)\times(-9)}=\sqrt{(-1)\times(4)\times(-1)\times(-9)}=\sqrt{-1}\times\sqrt{-1}\times\sqrt{4\times 9}=\pm i\times\pm i\times\pm 6 =\pm 6$$

What you did is like saying that sqr fails (in general) because

$$\sqrt 4=2\ne\sqrt 4=-2$$

Last fiddled with by LaurV on 2019-09-28 at 01:52

2019-09-28, 02:14   #5
retina
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Quote:
 Originally Posted by VBCurtis The square root of x^2 is |x|, not x.
Wot?

My teacher taught me that a square root has two answers, a negative and a positive.

sqrt(x²) = ±x

 2019-09-28, 02:30 #6 VBCurtis     "Curtis" Feb 2005 Riverside, CA 34·59 Posts In every precalc & calculus curriculum I've seen, square root is a function. That means it has one solution, the positive one. Consider a graph of y = sqrt(x); it's the top arm of a sideways parabola, not both arms. Contrast that with the action of taking the square root of both sides of an equation (say, to remove a power of 2 from a variable). That action leads to two solutions, a positive sqrt and a negative sqrt. x^2 = 9 has two solutions, but sqrt(9) = 3. Last fiddled with by VBCurtis on 2019-09-28 at 02:31
 2019-09-28, 13:42 #7 wildrabbitt   Jul 2014 3·149 Posts I knew the definition of root x as a function is the positive root but still I was bamboozled by the statement I wrote (I read it somewhere). Is the same true for $$x^4$$? -1^4 = |-1| = 1 $$\sqrt[4]{((-\sqrt{2})(-\sqrt{2})(-\sqrt{2})(-\sqrt{2}))}=|-\sqrt{2}| = \sqrt{2}$$ ?
 2019-09-28, 15:03 #8 Nick     Dec 2012 The Netherlands 32128 Posts If x is a real number, then $$x^4$$ is easier because $$x^2\geq 0$$ so $$\sqrt{x^4}=\sqrt{(x^2)^2}=|x^2|=x^2$$ and hence $$\sqrt[4]{x^4}=\sqrt{x^2}=|x|$$.
2019-09-28, 15:11   #9
retina
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Quote:
 Originally Posted by Nick If x is a real number, then $$x^4$$ is easier because $$x^2\geq 0$$ so $$\sqrt{x^4}=\sqrt{(x^2)^2}=|x^2|=x^2$$ and hence $$\sqrt[4]{x^4}=\sqrt{x^2}=|x|$$.
Hmm?

(-2)^4 = 16
(+2)^4 = 16

16^(1/4) = ±2

2019-09-28, 17:40   #10
xilman
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Quote:
 Originally Posted by retina Hmm? (-2)^4 = 16 (+2)^4 = 16 16^(1/4) = ±2
Hmm?

(2i)^4 = ((2i)^2)^2 = (-4)^2 = 16.
(-2i)^4 = ((-2i)^2)^2 = (-4)^2 = 16.

ITYM, 16^(1/4) = ±2, ±2i.

More generally, there are n distinct n-th roots of unity.

Ah, the OP is interested only in real roots.

Last fiddled with by xilman on 2019-09-28 at 17:43 Reason: Added final sentence.

 2019-09-29, 02:00 #11 Dr Sardonicus     Feb 2017 Nowhere 13×349 Posts The fundamental issue with "wrong square root" difficulties is, in order to define non-integer powers, you need a logarithm. With a logarithm in hand, you can say $a^{k}\;=\;exp{k\log(a)}$ for any complex number k. For roots of positive real numbers a, and real values of k, everything is fine -- you've got a real-valued logarithm, defined by an integral. You can not, of course, define a logarithm of 0. For nonzero complex numbers other than positive real numbers, though (including negative real numbers), defining a logarithm is a real problem. For one thing, the logarithm of any such number will have a nonzero imaginary part. But the real problem is, once you allow complex exponents, you are stuck with the fact that the exponential function is periodic, because $e^{2i\pi}\; =\; 1\text{.}$ so "the" logarithm is only determined up to an integer multiple of 2*pi*i. This difficulty is usually dealt with by making a "branch cut" emanating from 0, which excludes closed curves with 0 (the origin) inside them. Different integer multiples of 2*pi*i give different "branches" of the logarithm. And unless the imaginary part of every logarithm in sight is exactly zero, there is every chance that applying the usual "laws of logarithms/exponents/roots" will put you on a different "branch" of the logarithm than the one you started with. And, if you get on the wrong branch of the logarithm, you make a monkey of yourself!

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