 mersenneforum.org Can some explain this?
 Register FAQ Search Today's Posts Mark Forums Read  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 10010101010112 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
Romulan Interpreter

Jun 2011
Thailand

5×1,889 Posts 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
Undefined

"The unspeakable one"
Jun 2006
My evil lair

22×5×307 Posts 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 12AB16 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 $$\sqrt{((-\sqrt{2})(-\sqrt{2})(-\sqrt{2})(-\sqrt{2}))}=|-\sqrt{2}| = \sqrt{2}$$ ?   2019-09-28, 15:03 #8 Nick   Dec 2012 The Netherlands 2·33·31 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 $$\sqrt{x^4}=\sqrt{x^2}=|x|$$.   2019-09-28, 15:11   #9
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

22·5·307 Posts 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 $$\sqrt{x^4}=\sqrt{x^2}=|x|$$.
Hmm?

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

16^(1/4) = ±2   2019-09-28, 17:40   #10
xilman
Bamboozled!

"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across

1068210 Posts 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 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 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!   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post Nils Hardware 11 2012-07-21 18:06 davieddy Math 9 2009-11-07 07:42 petrw1 PrimeNet 8 2007-08-11 18:28 jasong jasong 5 2007-07-19 00:43 jasong Information & Answers 3 2006-09-12 02:25

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