Slightly OT: simple math dilemma « Next Oldest | Next Newest »

 ▼ Chuck Senior Member Posts: 320 Threads: 59 Joined: Dec 2006 04-12-2011, 10:46 PM I was asked by a colleague yesterday about real solutions to equations, i.e., sqrt(1-x)-sqrt(x-3)= 0 Question: is 2 a solution? and at what level of math should it be accepted as a solution. Most math texts will say it's not a solution: if y=sqrt(1-x) and y=sqrt(x-3) are graphed, they don't intersect in the real plane, yet the real number 2 is a solution (can't argue with that). So what constitutes a real solution. I claim you can't slide into the complex world and then back to real world; or since 2 isn't in the real domain of the functions, it can't be considered as a solution. One thing led to another, and it got me thinking about the value of (-1)^(2/6). At first glance I "want" it to be +1, since we have "even" powers, (sort of). On second glance it could be 1/2 + sqrt(3)/2 i, (the principal sixth root of a negative is complex; and then squared). But, on third glance, basic algebra says to reduce all rational exponents to lowest terms before evaluating, which gives us -1. Hmmmm. The TI-8X calculators give the third-glance result (-1) and produces a graph from -inf to +inf (the typical cube root function). Mathematica gives the second-glance result (complex value), and a graph from 0 to +inf accordingly. Back On-track, slightly: my HP-15 gives -1 (as expected). Haven't checked the others. my 28S gives the complex answer (as does my Casio) and graphs for only x>0. I've yet to find one that gives +1 (bummer,even powers and all). What do other calculators (hp or non-hp) give,and what "should" it be? I now can only presume TI's are for the less mathematically mature student, and HP's are for the more mathematically advanced. (Ouch!) But I wait your opinions. Enjoy. ▼ Eddie W. Shore Posting Freak Posts: 764 Threads: 118 Joined: Aug 2007 04-13-2011, 01:15 AM Anything that solves the equation is a solution IMHO. Just not always the desirable or sensible one. Frido Bohn Member Posts: 117 Threads: 16 Joined: Jul 2009 04-13-2011, 05:43 AM sqrt(1-x)-sqrt(x-3)= 0 can be transformed into: sqrt(1-x)=sqrt(x-3) thus, squaring both sides leads to: 1-x = x-3 and consequently to: 1+3 = 2x and to: x = 2 Is this too simplistic? Regards ▼ Dieter Senior Member Posts: 653 Threads: 26 Joined: Aug 2010 04-13-2011, 02:04 PM Quote: x = 2 Is this too simplistic? Nothing it soo simplistic as long as it's correct. ;-) But this solution isn't. You ran into an absolutely classic error here. Consider the original equation ``` sqrt(1-x) - sqrt(x-3) = 0 ``` For a real result this implies that both 1-x >= 0 and x-3 >=0. ``` 1 - x >= 0 => x =< 1 x - 3 >= 0 => x >= 3 ``` So x must to be not more than 1, and at the same time it has to be greater or equal to 3. In other words, a (real) solution is impossible here. Your result 2 as well as any other. What's wrong with your result? There are two things. First, you did not check the valid domain for a possible (real) result. Since the obtained value 2 does not fall in the domain for a valid result, it simply is no solution at all. Second, the step where both sides of the equation are squared is the weak point here. While "-4 = 4" is obviously wrong, squaring turns it to "16 = 16", which is correct. ;-) There are lots of other occasions that look tricker than initially expected: ``` (x^2 - 1) / (x - 1) = (x+1)(x-1) / (x - 1) ; x-1 cancels out = x + 1 ; really ?-) ``` Of course the first and third line are not equal. Try to plot both graphs and look at the point where x equals 1. ;-) Dieter ▼ Maarten Ambaum (Reading, UK) Junior Member Posts: 23 Threads: 2 Joined: Jan 2010 04-14-2011, 05:14 AM Dieter, I don;t understand your last example. The two expressions _are_ the same: the only difference would be that (x^2-1)/(x-1) appears to be undefined at x=1, but the limit of x -> 1 for (x^2-1)/(x-1) is perfectly defined (just write x=1+e, for small e, to find (x^2-1)/(x-1) = 2+e, and take the limit e-> 0) The confusion in all this discussion is whether certain factors can be undefined, such as 1/(x-1) in your example, while the full expression (x^2-1)/(x-1) is perfectly valid and has correct continuous limits at all points on the real axis (which is the case). Maarten Frido Bohn Member Posts: 117 Threads: 16 Joined: Jul 2009 04-14-2011, 02:00 PM Quote:` (x^2 - 1) / (x - 1) = (x+1)(x-1) / (x - 1) ; x-1 cancels out = x + 1 ; really ?-)`Well, for me is(x^2 - 1) / (x - 1) = x + 1 If you factor(x^2 - 1)you get(x+1)(x-1)and thus, the equation I quoted is an identity.If you try it the hard way with cancelling out(x-1) off (x+1)(x-1) / (x - 1) you getx+1.Then, we have the "challenging" equation(x^2 - 1) / (x - 1) = x + 1 Factoring (x^2 - 1) gets(x+1)(x-1) / (x-1) = x+1Cancelling again x-1 gets x+1 = x+1 and needless to say x = x I never tried to graph x=x, and neither the ancients greeks tried to solve a mathematical problem by means of graphing. They used geometry, and it worked fairly well. Perhaps, somebody can show me a geometric solution of this '(x^2 - 1) / (x - 1) = (x+1)(x-1) / (x - 1)' problem? Manolo Sobrino Member Posts: 79 Threads: 3 Joined: Jun 2010 04-13-2011, 10:30 PM f(z)= + sqrt(1-z) - sqrt(z-3) over C has a root in z = 2, it comes from the imaginary part of f(z), there's nothing weird. (I prefer this one, sqrt(1-z) - sqrt(z-3) = 0 i sqrt(z-1) = i sqrt(3-z) (z-1) = (3-z) z = 2 ) Cheers Edited: 13 Apr 2011, 10:36 p.m. Valentin Albillo Posting Freak Posts: 1,755 Threads: 112 Joined: Jan 2005 04-13-2011, 10:08 AM . Quote: I claim you can't slide into the complex world and then back to real world Disagree. Here you are, a classic counterexample: Best regards from V. ` ` Mike (Stgt) Posting Freak Posts: 858 Threads: 80 Joined: Feb 2009 04-13-2011, 12:17 PM Quote: I claim you can't slide into the complex world and then back to real world No, there are situations where you may only find _real_ solutions by using complex math. Ciao.....Mike ▼ Mike Morrow Posting Freak Posts: 758 Threads: 9 Joined: Jul 2007 04-13-2011, 05:05 PM I always liked i^i = exp(-PI/2). The result "seems" (at first) so non-intuitive. Probably the group that has the most day-to-day routine control of a process whose behaviour is described by real and imaginary complex variables is AC electric generating plant control room operators. They control their station's output of REAL load (MW) and REACTIVE load (MVAR), which can be represented by vectors in phase quadrature with each other. Unlike the MW loading, the MVAR loading can be grossly varied with no effect upon the fuel consumption of the machine driving the generator. Thus it even has a sort of "imaginary" aspect to it. :-) Thomas Klemm Senior Member Posts: 735 Threads: 34 Joined: May 2007 04-13-2011, 01:30 PM The confusion arises because you don't define the domain of the sqrt-function. If it should give only real results then the best you could use is: 0 <= x. But with this condition you would get: ```0 <= 1-x -> x <= 1 ``` ```0 <= x-3 -> 3 <= x ``` These two conditions are contradictory, thus no solution exists. However when you allow all real numbers as domain of the sqrt-function you will get complex numbers as intermediate results. There's no problem with the fact that the end-result is real. Conclusion: without specification of the domain, a function is not defined, thus the equation is meaningless. Cheers Thomas Marcus von Cube, Germany Posting Freak Posts: 3,283 Threads: 104 Joined: Jul 2005 04-13-2011, 02:56 PM The result of (-1)^(2/6) is defined as a number which, raised to the power of 6/2 (or 3), gives -1, hence one of the three possible values in the complex plane. Is there a point in rearranging the expression to ((-1)^2)^(1/6) ? Or to ((-1)^(1/6))^2. The former would reduce the number of solutions to 1, the latter to 6 (which might collapse to less by squaring, I haven't checked.) In the latter case, 1 should be one of the solutions. Writing this down, I tend to the idea that the expression has six solutions in the complex plane, the real values -1 and 1 being amongst them. I'm still puzzled. ▼ Chuck Senior Member Posts: 320 Threads: 59 Joined: Dec 2006 04-13-2011, 11:22 PM They actually do collapse to only three values; two complex and -1 (using De Moivre's formula) . My desired +1 just isn't there. :( Chuck Senior Member Posts: 320 Threads: 59 Joined: Dec 2006 04-13-2011, 11:13 PM Excellent discussion. It seems the level of math a student has is what may determine the expected result. I presume a student who has not been introduced to complex numbers would have to assume square-roots of negatives are not possible and therefore no solution. A student with a rudimentary knowledge of complex numbers would conclude that 2 is an easily found and appropriate answer. And a student with a fairly thorough knowledge of complex numbers would recognize that the complex result for (-1)^(2/6) is preferred, but that there are a total of three solutions. Along with this, I find it very interesting the results that are given by various calculators and math programs. One can start to glimpse the mathematical sophistication of the manufacturers target consumer population. Fun stuff. Edited: 13 Apr 2011, 11:20 p.m. ▼ Manolo Sobrino Member Posts: 79 Threads: 3 Joined: Jun 2010 04-13-2011, 11:44 PM I think you're mixing things up a little bit, it's actually complex analysis 101 to prove that any complex number has n different nth roots, and the first problem's answer is x=2, rudimentary or not ;) Edited: 13 Apr 2011, 11:47 p.m.

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