Squaring and Square roots

[sebbysteiny]

Moderator note: The purpose of the projects and homework forum is to assist students with school or self assigned homework. It’s inappropriate to introduce mathematically unconventional claims in this thread, so I’ve moved the first 2 posts in this thread from project and homework to Mathematic and Physics.

- CraigD

 

Just a little to add.

 

You may need to give a little thought to +/- signs when squaring both sides. This is particularly so for taking the square roots.

 

The problem is that sqr (x^2) = +x or -x.

 

So if you take the square root of both sides, you lose solutions if you just take the positive answer. If you square both sides, you may gain solutions.

 

Here, the solution to [sqr ((x+y)/(x-y)]^2 = 1/4 that gives y = -3/5 x

 

may be a solution to either sqr ((x+y)/(x-y)) = 1/2, as given in the question,

 

OR to sqr ((x+y)/(x-y)) = -1/2, which was not the equation given in the question.

 

It seems very difficult to tell which one you have actually solved.

 

So you now need to show that y=-3/5 x is the solution to the original equation and not the corresponding one artificially obtained from squaring both sides.

 

Also, since you have now squared the thing, you should have two solutions. Where and what is the other? Somewhere along the line, you have lost a solution.

 

My advice on this is not to square or sqare root both sides of the equation unless it's absolutely necessary. Otherwise you have the awkward job of having to follow both possible solutions at every stage to see which equation you are actually solving.

[sebbysteiny]

Having thought about this further, the problem seems to be not in your working, but in the original question.

 

sqr ((x+y)/(x-y)) = 1/2 cannot possibly be correct, can it?

 

Surely, for all values of real x and y, sqr (f(x,y)) [ie any function of x and y] = c [a constant], that contant must be both +ve or -ve.

 

So the problem seems to me to be that the teacher created the orignal problem backwards who, in square rooting both sides to make the original problem, forgot about the -ve solution, which is kind of necessary information in my view.

 

Okay, enough postulating. This has got me so wound up I am going to do something I have not done in 2 years: actually plug numbers into equations. Mathematics: I'm back.

 

sqr((x-3/5x)/(x+3/5x)) = 1/2.

 

sqr((x([2/5] / x[8/5)) = 1/2

 

sqr((x/x * 1/4)) = 1/2

 

sqr(1/4) = 1/2 unless x = 0.

 

+/- (1/2) = 1/2 unless x = 0.

 

So there we have it. Finally and conclusively. I believe the question is invalid. y=-3/5x cannot be a solution as, if it were, it would lead to both + 1/2 and - 1/2 answers.

 

Also, y is not equal to -3/5x if x = 0.

 

Sloppy question.

[CraigD]

sqr ((x+y)/(x-y)) = 1/2 cannot possibly be correct, can it?

Yes, I believe it can. Through valid algebraic transformations, it can be show to be equivalent to

 

[LATEX]y = -\left(\frac35\right)x[/LATEX].

Surely, for all values of real x and y, sqr (f(x,y)) [ie any function of x and y] = c [a constant], that contant must be both +ve or -ve.

This isn’t the conventional definition of the square root function. If it were (call the function “altsqr()” and define it altsqr(x) = +/- sqr(x), or altsqr(x) = +/- sqr(x)), they sebbysteiny’s claim would be true. That is a real number (though not the complex number) espression

-sqr(x) = 1/2

, is false.

Also, y is not equal to -3/5x if x = 0.

True.

 

Because the set of real numbers is not closed under the operation of division, expressions containing division may not be closed, so neither

sqr ((x+y)/(x-y)) = 1/2

nor

(x+y)/(x-y) = 1/4

are closed for all real x and y. When x = y = 0, [math]\frac{0+0}{0-0} \not= \frac14[/math]. It is undetermined.

 

Note that

(x+y)/(x-y) = 1/4

can be rewritten

4(x+y) = (x-y)

, in which case it is determined for x = y = 0. By eliminating the division operation, the expression becomes closed for the set of real numbers.

[Qfwfq]

Sorry Craig, but he isn't introducing anything mathematically unconventional. The wiki you link to is about the square root, which is relevant to equations of this type, and isn't about the square root function. It even says: "The other square root of 9 (not the principal square root) is −3." I do agree that the equation

 

[math]\sqrt{\frac{x+y}{x-y}}=\frac{1}{2}[/math]

 

has solutions, it just isn't a good example.

[sebbysteiny]

Hmm, I guess, on reflection, that might have been slightly advanced :cocktail:.

 

Still, I think that students be aware of the implications of squaring and squarerooting both sides of an equation in terms of the signage and the number of solutions. Although I agree this should not extend to a complete and accurate use of signage in their working out until a higher level.

 

The question in the original thread that, if you pardon the expression, got my tits up was whether the equation sqr((x+y)/(x-y)) = 1/2 contained all the necessary information even allowing for the definition of sqr as being the positive square root.

 

It seemed clear to me that, unless you give careful treatment to the signage, if you square root each side of the equation, you lose solutions. If you square it, you should gain one.

 

But the squared version only had one solution.

 

Therefore, the original equation should have no solutions.

 

So what is going on here????? You can't just take square roots and sqares of both sides without considering the implications on signage.

 

Mathematically, did you also need the knowledge that

-sqr((x+y)/(x-y)) = 1/2

to find a solution ? Or is the original formultion sufficient? And if it is complete, then where is the extra solution one would expect from squaring both sides?

 

True.

 

Because the set of real numbers is not closed under the operation of division, expressions containing division may not be closed, so neither

sqr ((x+y)/(x-y)) = 1/2

nor

(x+y)/(x-y) = 1/4

are closed for all real x and y. When x = y = 0, . It is undetermined.

 

I'm afraid I don't agree with you here. The only reason why the reasoning I used did not hold for x=0 was because I called x/x = 1, which is true for x = 0.

 

Yes, it so happens, that at x = 0, x-y = 0 too making the equation undefined, but just because the equation is undefined at 0, I do not believe that means the relationship between x and y does not hold at zero.

 

eg (x (2y +x) / (x (2y - x +1)) = (2y + x)/(2y - x + 1) unless x = 0 even though the expression is now defined at x = 0.

 

But I'm sure you know this already.

 

You are quite right, though, my working could have potentially avoided this by multiplying both sides by sqr (x) rather than canelling out the x on top with the x on the bottom.

 

Lets see if it works.

 

sqr((x-3/5x)/(x+3/5x)) = 1/2.

 

sqr((x([2/5] / x[8/5)) = 1/2

 

sqr((x/x * 1/4)) = 1/2

 

(1/sqr (x)) * sqr (x/4) = 1/2.

 

sqr (x/4) = sqr (x) / 2.

 

sqr (x/4) = sqr (x/4).

 

Fair enough. It holds even if x = 0.

 

QP for the latter spot.

 

If you can resolve the problem with signage too, I'll give another QP.

[CraigD]

The only reason why the reasoning I used did not hold for x=0 was because I called x/x = 1, which is true for x = 0.

Reasoning such thing as [math]\frac00=1[/math], [math]\frac{2a}0 -\frac a0 = a[/math], etc. leads to a famous mathematical fallacy that can be used to “prove” contradictions such as [math]1=2[/math]. This wikipedia article section has a example and a brief discussion.

 

The fallacy points to a deep “philosophical” (apologies to sebby for using the term ;)) dichotomy in math that’s at least 30 years old*, between the position, which could roughly be termed “idealist”, and another I’ll call “nomenclaturalist”. One position holds that math is a shorthand for describing “real” things if not actual physical reality, then Plato’s ideal forms. The other holds that math is a collection of “nomenclatural rules” about manipulating the symbols used to write math. The first position is responsible for asserting that [math]0/0=1[/math] is a fallacy. The second would hold that that it’s a perfectly legitimate application of a formal system in which [math]a/a=1[/math] without exception, but would point out that [math]1=2[/math] is a theorem (true statement) in this formal system, making it a weird, counterintuitive one.

 

* The dichotomy is at least 30 years old, because I personally encountered discussion of it in Hofstadter’s Gödel, Escher, Bach. GEB is, IMHO, essential reading for Math enthusiast and professionals. I’m appalled on occasions when I encounter academic mathematicians who’re unacquainted with it, experiencing a sense of dissonance as if meeting a rabbi who’s never heard of the Kabbalah.

[Qfwfq]

Alright fellas, it seems you both have some details right and others wrong IMESHO.

 

It seemed clear to me that, unless you give careful treatment to the signage, if you square root each side of the equation, you lose solutions. If you square it, you should gain one.

Not always. For a given equation, squaring it MAY give an equation with extra solutions.

 

Also, y is not equal to -3/5x if x = 0.

But at the same time, it couldn't be other than 0 (which equals -3/5 times 0) because, with x = 0, the equation becomes

 

[math]\frac{y}{-y}=-\frac{y}{y}=\frac14[/math]

 

and for any value of y other than 0 this is -1 = 1/4.

 

True.

 

Because the set of real numbers is not closed under the operation of division, expressions containing division may not be closed, so neither

sqr ((x+y)/(x-y)) = 1/2

nor

(x+y)/(x-y) = 1/4

are closed for all real x and y.

Using only algebra. It might not make sense to say what an expression is for x = a but it can make sense to say what [math]\lim_{x\rightarrow a}[/math] of it is.

 

When x = y = 0, [math]\frac{0+0}{0-0} \not= \frac14[/math]. It is undetermined.

Actually, it makes sense neither to say it is [math]\norm =\frac14[/math] nor that it is[math]\norm\not=\frac14[/math] Undetermined doesn't mean that it can't be some given value. It does make sense though, with y = -3x/5, to say that

 

[math]\lim_{x\rightarrow 0}\frac{x-\frac35x}{x+\frac35x}=\frac14\lim_{x\rightarrow 0}\frac{x}{x} =\frac14[/math]

 

So, with the condition x = 0 y is constrained to be 0 which is compatible with y = -3x/5 and this relation is valid for all values of x.

 

;)

[sebbysteiny]

The fallacy points to a deep “philosophical” (apologies to

sebby for using the term ) dichotomy

 

That's okay.

 

About the minor issue of dividing by zero.

 

The point I was saying is that, since my argument put x/x = 1, it is not valid at x=0, my proof only works when x not = 0.

 

You however pointed out a better proof which proved the relationship holds even when x = 0.

 

However, you cannot say simply because the equation has a denominator of x-y and y =-3/5x, that the relationship does not hold. For starters, it can be calculated using 'other ways', such as limits, to get around the 0/0 problem.

 

I'm guessing we all agree here.

 

But now for the more major problem of the squaring and square rooting of the equation.

 

It is fundamentally impossible to square both sides of an equation without creating more solutions.

 

(1) F(x,y) = G(x,y).

(2) F(x,y)^2 = G(x,y)^2.

 

So we have now squared both sides of a general equation with two variables.

 

But, now,

 

(3) F(x,y) = - G(x,y) is also a solution.

 

So equation 2 will produce solutions of both equation 1 (the original) and equation 2 (which we do not want) and it is tricky to tell exactly which solution to 2 is a solution to 1 rather than 3.

 

So if sqr((x+y)/(x-y)) = 1/2 has solutions,

 

squaring both sides will give both solutions to sqr((x+y)/(x-y)) = 1/2 and -sqr((x+y)/(x-y)) = 1/2

 

So the question I've been getting severely irritated with all morning is....

 

are those solutions the same? If so, why? If not, where are they?

 

I'm going to now play around with the algebra and see what I get as I have a hunch.

 

sqr((x+y)/(x-y)) = -1/2

 

sqr((x+y)/(x-y)) + sqr(1/4) = 0.

 

(x+y)^1/2 + [(x-y)^1/2 ]/2 = 0.

 

2/5x^1/2 + 2/5x^1/2 = 0. :eek:

 

This cannot be. According to this y = -3/5x cannot be a solution to sqr((x+y)/(x-y)) = -1/2.

 

The corresponding equation if +sqr((x+y)/(x-y)) = 1/2 gives

 

2/5x^1/2 - 2/5x^1/2 = 0. This is true, so y=-3/5x is a solution to the actual question.

 

This suggests two thing to me: either there are no solutions to

-sqr((x+y)/(x-y)) = 1/2, real or imaginary,

or there has to be a second solution to the equation ((x+y)/(x-y)) = 1/4.

 

This just keeps getting more confusing

 

I'll give a QP to anybody who can solve this dilema [unless, of course, for some reason, I have had a big personal dispute with them]

[CraigD]

When x = y = 0, [math]\frac{0+0}{0-0} \not= \frac14[/math]. It is undetermined.

 

Actually, it makes sense neither to say it is [math]=\frac14[/math] nor that it is[math]\not=\frac14[/math] Undetermined doesn't mean that it can't be some given value.

Mea culpa – I stand corrected. I should have used “indeterminant” rather than “undetermined”.

 

Both terms are, entomologically, “procedural” – they are defined by whether a particular mathematical variable is known, or knowable, from the expressions containing it. These definitions emerge from the “nomenclaturalist” interpretation of Math.

 

Under an “idealist” interpretation, indeterminate means “not an element of the set”. It’s unusual to use the term in this context - “not closed” is more common.

 

 

Because the set of real numbers is not closed under the operation of division, expressions containing division may not be closed, so neither

sqr ((x+y)/(x-y)) = 1/2

nor

(x+y)/(x-y) = 1/4

are closed for all real x and y.

Using only algebra. It might not make sense to say what an expression is for x = a but it can make sense to say what [math]\lim_{x\rightarrow a}[/math] of it is.

…

It does make sense though, with y = -3x/5, to say that

 

[math]\lim_{x\rightarrow 0}\frac{x-\frac35x}{x+\frac35x}=\frac14\lim_{x\rightarrow 0}\frac{x}{x} =\frac14[/math]

 

So, with the condition x = 0 y is constrained to be 0 which is compatible with y = -3x/5 and this relation is valid for all values of x.

True.

 

Limits, and calculus, can be considered algebra with a peculiar number type – the infinitesimals, to use a more modern (post 1960) term, hyperreal numbers. There’s a good bit of evidence that Newton and Leibniz though about them, but so uncomfortable with them and the whole of number theory that they chose to put the calculus in terms of the limits of sequences of complex and real numbers, a formalism that remains the dominant convention - or at least was when I last had a class in it, in the 1980s. Newton, the infinitesimals, and one of their recent influential mathematicians, Abraham Robinson were recurring topics of discussion among my class of Math undergrads, for whom the “everything’s a set” brand of idealism was the dominant paradigm.

[sebbysteiny]

Cool. I think we have all now agreed on the limits at 0 thing.

 

What we have yet to answer is what was going on when we squared both sides and the effect of the -ve root solution.

[LJP07]

Glad my question posed some debate, but as regards my other thread, can ye solve problem two?

[Qfwfq]

Yeah, infinitesimals as defined in terms of limits are still the standard. Hyperreal numbers are a pioneering thing, no good concrete construction has yet been found.

 

What we have yet to answer is what was going on when we squared both sides and the effect of the -ve root solution.

Essentially, the set of solutions of the equation:

 

[math]a^2=b^2[/math]

 

is the union of those of both:

 

[math]a=b[/math] (1) and:

 

[math]a=-b[/math] (2)

 

Of course, when one takes the root of each side of a given equation, taking the principle root of both sides gives (1) rather than (2). I can still remember when my lousy high-school teacher just said that squaring both sides could "introduce extraneous solutions" so we must verify each solution that we find, for the original equation, with no better explanation. Not hard to figure out, but it isn't always so easy for a teen-ager to get it clear on their own and I felt somewhat frustrated.

 

:cup:

[CraigD]

I’ve put an embarrassing :doh: amount of effort into solving the equation Sebby gave in post #8:

 

[math]\sqrt{\frac{x+y}{x-y}} = -\frac12[/math]

 

For complex number values of [math]x[/math] and [math]y[/math].

 

Assuming the definition [math]i = \sqrt{-1}[/math], this begins

 

[math]\frac{x+y}{x-y} = \frac{i}4[/math],

 

and ends with the pair of equations

 

[math]4a +b +4c -d = 0[/math]

[math]-a +4b +c +4d = 0[/math]

 

where [math]x=a+bi[/math] and [math]y=c+di[/math].

 

So any [math]x[/math] and [math]y[/math] that satisfy the pair of equations and the constraint that ([math]a \not= 0[/math] or [math]b \not= 0[/math]) and ([math]c \not= 0[/math] or [math]d \not= 0[/math]) and [math]x \not= y[/math] is a solution, example: [math]x=17 , y= -15 +8i[/math]; [math]x=17 +34i , y=-31 +22i[/math].

 

I was surprised how much more complicated a complex solution with imaginary parts is than the real number solution. :xx:

[sebbysteiny]

I’ve put an embarrassing :xx: amount of effort into solving the equation Sebby gave in post #8:

 

...

 

I was surprised how much more complicated a complex solution with imaginary parts is than the real number solution.

 

I was going to post just 'genius, simply genius' and then give you the biggest QP I've ever given (which, admittidly is no bigger than all the others), but I saw something out of the corner of my eye that may amount to the tiniest of technicalities.

 

I don't want you to hate me CraigD.

 

But I just want to say that....

 

please don't hate me CraigD,

 

the only slight question I want to make is ...

 

remember, you promised not to hate me, and it is illogical to attack the messanger,

 

If sqr((x+y)/(x-y)) = -1/2 = i^2/2,

 

isn't (x+y)/(x-y) = i^4/4 = 1/4 rather than i/4? :friday:

 

Please don't hate me.

[Qfwfq]

I'd say the messenger is right, and shouldn't be hated. ;)

 

[math]\sqrt{\frac{x+y}{x-y}}=-\frac12[/math] is just the (2) of my post above, in this specific case.

[CraigD]

If sqr((x+y)/(x-y)) = -1/2 = i^2/2,

 

isn't (x+y)/(x-y) = i^4/4 = 1/4 rather than i/4?

Yes, you’re right, thank you. I actually solved

 

[math]\left( \frac{x+y}{x-y}\right)^2 = -\frac1{16}[/math]

 

It was a good exercise in algebra of complex numbers, and brought back stuff I’ve either long forgotten or never known, but not the problem I intended. The moral of the sad tale is, I think, don’t be hasty in your first step!

 

I’ll get back to the original when I have some time to spare. Non-integer exponentiation is a whole different beast than simple algebra, so requires a major shift in mental gears, which I won’t have time for for at least 72 hours. Fun stuff.

[sebbysteiny]

I thought I might just add a post to bring this thread up to the top because this question is still bugging the hell out of me.

 

Does the equation using -ve roots have any solutions at all, real or imaginary?