# Getting i^3 = -i using laws of surds

1. ### E_Q

15
When doing FP1 ages ago I discovered a way to seemingly get i3=±i, and since then nobody has provided a proper explanation for why the positive solution is impossible.

Standard solution
i3=i2.i
=-1.i
=-i

'flawed' solution
i3=(√-1)3
=√(-1)3 Using the laws of surds, √a.√b=√ab
=±√-1
=±i

Of course, this applies to i2 just as well:

i2=√-1.√-1
=±√1
=±1

I'm not really suggesting the method is flawed, just is gives an incorrect answer as well as the correct one. I'm familiar with this (from solving modulus graphs, for example), but can't understand why the positive solution is incorrect in this case.

Thanks for any help!

### Staff: Mentor

It's because in taking the square root there are always two solutions

The sqrt of 4 is 2 and is -2 as both are 4 when squared.

In your case you have chosen only one value and discarded the other.

When working with complex numbers multiplying by i means a rotation of 90 degrees in the counterclockwise direction hence the reason you get -i as the answer.

### Staff: Mentor

You're missing the fine print on this property: Both a and b have to be nonnegative.

### Staff: Mentor

Assuming that x ≥ 0, the notation √x represents the principal, or positive square root of x.

Although it's true that each positive real number x has two square roots, which might be what you're saying below, √x represents the positive one of these roots.

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5. ### E_Q

15
I discared neither value, that's why I left the ± notation. This is the answer I wanted though; i3 represents a transformation of the point (1,0) 270° anticlockwise (like you said multiplying by i rotates 90° ACW, and i3 is equal to 1.i3), which of course becomes -i on an argand diagram.

Obvious once you think of it. Thanks!

In fact, that's a far easier way of working out in...just picture an argand diagram and think where the point would be transformed to!

### Staff: Mentor

Which is incorrect, for two reasons.

First, you misapplied the property that ##\sqrt{a}\sqrt{b} = \sqrt{ab}##. Second, you said that ##\sqrt{1} = \pm1##.

In your work you concluded that i2 = ±1, which is incorrect for the reasons listed above.

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7. ### E_Q

15
Why do the rules of surds not apply for negative numbers? I've never head this before. And it appears to hold true, it simply provides an extra solution which is incorrect, as well as the correct solution.

I'm similarly confused on your next point; √1=±1... Take the inverse, (-1)2=1 and 12=1

"All positive real numbers has two square roots, one positive square root and one negative square root."

http://www.mathplanet.com/education/algebra-1/exploring-real-numbers/square-roots
(Lame reference, I know).

But, thinking of a graph y=√x, there can be no real solutions for a negative x, obviously, not least as the range of y=x2 graph is ≥0, and as that is the inverse of y=√x its domain must also be positive numbers. But I would have imagined the graph might reflect and be repeated in the x axis to allow for both negative and positive solutions.

If you want this thread to progress at all you have to explain yourself, please. I am aware that in both my above examples, i3 and i2 there is only one answer, and sought to understand the reason the positive and negative answers, respectively, were wrong.

### Staff: Mentor

Because negative numbers inside the radicals give incorrect results..
Your equation √1 = ±1 is incorrect. By definition, √1 = 1. When you say "Take the inverse" what you mean is "square both sides."

The trouble with squaring both sides of an equation is that it is not a reversible operation. If you square -1, you get 1, and if you square +1, you also get 1. Going the other way, you can't get two results out of taking the square root of 1.

From a geometric perspective, the function f(x) = x2 is not a one-to-one function, so it doesn't have an inverse that is a function. If you restrict the domain of f to, say x ≥ 0, then you now have a one to one function. Given a value (≥ 0), you can find its square. Given a y-value, you can find the x-value that produces that y-value.
Aside from the grammar, which should be "All positive real numbers have ...", the sentence is correct. We denote them as √x and -√x.
I talked about this above. If you have y = x2, and x≥ 0, the graph is the right half of the parabola y = x2. In this case, these two equations are equivalent:
y = x2, x ≥ 0
x = √y

Given an x value, you can use the first equation to find the y value. Given an x value, you can use the second equation to find the y value. Notice that since the equations are equivalent, they have exactly the same graph, which is the right half of the parabola.

If you swap x for y in the second equation, this causes the graph to be reflected across the line y = x.

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### Staff: Mentor

Something that is a pet peeve of mine is the way function inverses are often taught at the precalc level. Given a function y = f(x), the steps are usually presented like this.
1. Switch x and y in the equation y = f(x).
2. Solve for y to get y = f-1(x).
If you graph both equations, you should see that each is the reflection of the other across the line y = x.

IMHO, step 1 is unnecessary and the whole procedure be rewritten as
1. Solve the equation y = f(x) for x.
2. Stop.
Solving the equation y = f(x) for x, gives you x = f-1(y).

The reasons that I dislike the first procedure above are that 1) a fair number of students quit after swapping x and y and think they are done, 2) the procedure obscures the fundamental relationships between a function and its inverse, and 3) the whole "swapping" process is completely irrelevant to how inverses are actually used in later courses such as calculus, differential equations, linear algebra, etc.

The whole idea is that if you are working with a one-to-one function, say y = f(x) = x3 + 3, this formula produces the y-value on the graph if you know an x-value. If you want to go the other way, you need the inverse.

Here's an example problem: Find the point on the graph of y = f(x) = x3 + 3 that intersections the line y = 11. To do this, we need the inverse, namely f-1(11). Here's the work.

y = x3 + 3
##\Rightarrow## y - 3 = x3
##\Rightarrow \sqrt[3]{y - 3} = x## (= f-1(y))
And we're done. From this formula, we see that f-1(11) = (11 - 3)1/3 = 81/3 = 2.
The graph of y = x3 + 3 crosses the line y = 11 at the point (2, 11).

Notice that f(2) = 23 + 3 = 11, and that f-1(11) = 2, as already calculated.

Another example that really shows how silly the whole idea of swapping variables is the conversion formula between Celsius and Fahrenheit temperatures.

F = f(C) = (9/5)C + 32.

If you're asked to find the inverse relationship (converting from F to C, it makes no sense at all to swap the variables. If you do that, you get a mess.

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10. ### E_Q

15
Quite right. Loose talk, not mathematical.

Yes I did think of this, but didn't realise that it meant the graph was invalid without saying x ≥ 0. Correct me if wrong: y=x2 is a one-to-many function, which is allowable, but this means it cannot have a function inverse as that would be many-to-one (domain has multiple solutions), the x2 parabola reflected in y=x, as you said.

So all those parabolas I used in FP1 were not classed as functions?

I am familiar with working out inverses and was taught your peeve way (swap y and x); obviously it's wrong to not make the final step and say ...=f-1(x). It's a bit irrelevant although I agree that it's more convenient to work out inverses in terms of y, as this is what you would do intuitively anyway. y=x-5, you create the inverse by saying y+5=x, so f-1(y)=y+5. You wouldn't work out f-1(x) by doing x=y-5, x+5=y. I imagine the reason it's taught like that is so students are reminded that the domain of the function becomes the range of the inverse and vice versa.

Actually when I saw your example I just thought:
y = x3 + 3 intersects y=11
∴solutions at x3+3=11
x3=8
x=2 (or should that be ±2...?)
Definitely a simpler way. But maybe you were trying to illustrate the use of inverses.

I noted the mistake but after all you can't edit a quote, can you...:grumpy: This seems odd though. If you say that every positive real number has both a positive and negative square root, how do you denote square rooting?

You said √1=±1 is incorrect, so how should it should be written? SQRT(1)=±√1=±1?
I think this is a very pedantic use of notation...

Thanks for the input though, it's made me think a little deeper about functions and inverses (it's very easy to mechanically swap y's and x's...)

### Staff: Mentor

x2=1 has two solutions, 1 and -1, or for short, ±1.

No, it's not. It's a very important point. √x is a function that maps [0,∞) to [0,∞) such that (√x)2=x for all x in [0,∞). There's no problem here with the fact that (√x)2 is also equal x because the definition of the function arbitrarily excludes that other solution.

This function can be extended to ℂ→ℂ, but now the fact that there are two solutions to z2=c is inescapably going to raise problems. There's a branch point at the origin. You have to pick a line segment (or more generally, a curve) that emanates from the origin such that √z switches from one family of solutions to the other across that line. Generally that line is taken to be the negative real axis.

### Staff: Mentor

I didn't say anything about the graph being invalid. All I said was that y = x2 is not one-to-one, so it doesn't have an inverse.
Yes, you are wrong. It's not one-to-many - it's many-to-one, or more accurately, two-to-one. For each x, both x and its opposite (-x) are paired to a given y in the range.

A one-to-many relation is not a function. There's a very simple test that you can use to tell whether a graph is that of a function - if a vertical line ever intersects with more than one point, the graph is not the graph of a function.
Correct. If you reflect the graph of y = x2 across the line y = x, you get a graph whose equation is x = y2, a parabola that opens to the right. This graph does not represent a function, as each x value is paired with two y values.
I don't know what FP1 is, so I can't answer your question.
But my whole point is that swapping x and y (and therefore reflecting across the line y = x) is silly. More importantly, that's not what you do in subsequent math courses. To show how silly this process is, see what you get by swapping variables for the C-to-F conversion formula F = (9/5)C + 32.
I'm hopeful that you're kidding...
Yes, I was. In symbols, here's what you are doing:
f(x) = 11
==> f-1(f(x)) = f-1(11)
The left side of the equation above simplifies to x. Using my formula of f-1(y) = ##\sqrt[3]{y - 3}##, we get that x = 2. (And definitely NOT ±2!)
I can, but then again, I have super powers. :tongue:
When you take the square root of something, you get the principal, or positive, square root.
√1 = 1 Period.
And √4 = 2, √9 = 3, √16 = 4, √25 = 5, etc.
Or using your notation, SQRT(1) = 1.
When you enter a number in a calculator and press the √, it doesn't give you two values does it? When you look at the graph of y = √x, there's only one y value for each x in the domain, right?
I disagree. It's not pedantic if the alternative is wrong.
... and much harder to actually understand what you're doing. This is why the usual way of presenting inverses seems so inane to me.

13. ### E_Q

15
Think I've grasped the idea now.
x2=4
x=±√4
x=±2

But if you are simply presented with
x=√4
then the only solution is
x=2
as it refers only to the positive root. It's a mistake that often doesn't get picked up on that √4=±2, as if you then square both sides you lose the negative anyway. Thanks for clearing this up (@Mark44 also)

Unfortunately this paragraph has gone almost entirely over my head as I don't know any of the notation here apart from 'z'. c, [0,∞), ℂ (same as c?)

Pretty annoyed I made that mistake. At least I asked to be corrected; perhaps it's some instinct that I was being stupid.

In FP1 (further pure 1, Edexcel) there was a chapter on parabolas in the form y2=4ax where (a,0) is the focus. This would be a one-to-many function (as there are two y solutions for each x); so does not qualify as a function?

But its inverse:
y2=4ax
y2/(4a)=x=f-1(y)
would be a valid function, being a one-to-many function?

F = (9/5)C + 32
F - 32 = (9/5)C
(F - 32)(5/9) = C = f-1(F)
Correct...

F = (9/5)C + 32
C = (9/5)F + 32
(C - 32)(5/9) = F = f-1(C)
Utterly useless, as far as I can see. Does this have any point whatsoever? I think I may understand your view, now...

### Staff: Mentor

I've never heard of this.
There is no such animal. A relation can be one-to-many, but such a relation is NOT a function.
When you say "valid" function, the implication is that some functions are valid, and others are invalid. A relation (which is a pairing between objects in one set and objects in the same or another set), is either a function or it isn't. If the graph of the relation has any point that is directly above another, the relation is not a function. This is called the Vertical Line Test.

Solving for x, you get ##x = \frac 1 {4a} y^2##. Here x IS a function of y, albeit one that is not one-to-one. You could call it a many-to-one function where "many" here means "two". There's another test, the Horizontal Line Test, that can be used to determine whether a function is one-to-one or not. If a horizontal line ever intersects more than one point on the function's graph, the function is not one-to-one.

15. ### E_Q

15
grrr I'd love to use the phrase "you know what I mean" but I'm sure you'd argue what I meant was wrong :tongue:. You're right, but I'd not heard of the term "relation" before, so I just stuck with function ^.^

The loose use of the word function is because that's how my maths teacher says it..."f(x) is not a valid function because it's a one to many function"

I'll make sure to remind my friends of the common sense *ahem* horizontal line test when they're working out what sort of function a relation is, if at all.

Well, I think I've got my terminology set :D Funny how this topic started off with i...

### Staff: Mentor

It is *not* a mistake. It is very intentional. √4=2, not ±2. √x is a function. Something that maps a value to many values is not a function. Some call these things that map one to many a multivalued function. That's a bit of a misnomer because a multivalued function is not a function.

Why all this harping about functions? Functions bring a lot of power to the table that is lost when the mapping is to a set of values rather than to a single value. Functions are a central (perhaps *the* central) concept in mathematics.

ℂ (better: ##\mathbb C##) denotes the set of all complex numbers. Similarly, ℝ (better: ##\mathbb R##) denotes the set of all real numbers.

17. ### Tobias Funke

159
I see a lot of surprise and sometimes flat out resistance to the fact that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ doesn't always hold (not saying you're resistant, but there were a couple of recent threads where someone simply wouldn't accept this fact; so much so that I was shocked they weren't locked almost immediately). Is it really so surprising? If you're talking to a four year old and you tell them that $x$ is bigger than $y$, then it most likely follows that $x-y\geq 1$. But that certainly isn't true for a 15 year old. Rules that always worked before don't always work in new situations.

I also wonder how much of this is due to teaching that $i=\sqrt{-1}$ (instead of $i^2=-1$) without explaining how to interpret what that means. The only thing most students can deduce from that is that $i\geq 0$, which makes no sense, but is completely justified on their part. And your flawed solution does indeed make the substitution $i=\sqrt{-1}$.

### Staff: Mentor

You're not really "solving" an equation here, as √4 and 2 are just different names for the same number.
I'll go out on a limb here and posit that what E_Q really meant but didn't get across clearly was this, "It's a mistake that √4=±2, that often doesn't get picked up on..."

19. ### AllyScientific

13
I think Tobias is talking about my recent threads where I showed my resistance to the supposed fact that $\sqrt{ab} = \sqrt{a} \sqrt{b}$ doesn't always hold.

If it makes no sense that $i$ is positive in $i^2$=-1, then $i$ is negative.
But does it make sense that a negative number $\times$ a negative number is equal
to a negative number ?

### Staff: Mentor

The problem here is that you are applying Real number notions that you learned to an imaginary number i and by extension to the complex number plane where i can be +i or -i.