- #1

- 24

- 0

a simple question

what is zero "0" does it count as a number?

any answers

what is zero "0" does it count as a number?

any answers

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter hamlet69
- Start date

- #1

- 24

- 0

a simple question

what is zero "0" does it count as a number?

any answers

what is zero "0" does it count as a number?

any answers

- #2

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 967

"0" is the additive identity (that's another).

Yes, 0 is a number just like -1, π, √(2) and i.

- #3

- 28

- 0

0 is the mark i will get on my linear exam if i don't study ;)

- #4

- 248

- 0

Also, zero is both the limit of the largest negative numbers and the limit of the smallest positive numbers...

Naturally, even more important (due to Euler):

[tex] 0 \; = \; e^{i\pi}-1[/tex]

Had enough?

- #5

- 650

- 1

I dont know much of its importance given by Our Ancient Seers

But mathematically it is an essence in every field

Metaphysically it represents DEATH,GLOOMY,INAUSPICIOUS

- #6

jcsd

Science Advisor

Gold Member

- 2,097

- 12

Zero is the identity element in addition (of vectors, integers, etc.)

edited to add: looks like HallsofIvy has beaten me to the punch on that defintion.

edited to add: looks like HallsofIvy has beaten me to the punch on that defintion.

Last edited:

- #7

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 967

Only by 19 days!

- #8

- 16

- 0

zero makes mathematics full of identities and definitions....

- #9

- 100

- 0

That should beOriginally posted by suyver

Also, zero is both the limit of the largest negative numbers and the limit of the smallest positive numbers...

Naturally, even more important (due to Euler):

[tex] 0 \; = \; e^{i\pi}-1[/tex]

Had enough?

[tex] 0 \; = \; e^{i\pi}+1[/tex]

- #10

matt grime

Science Advisor

Homework Helper

- 9,420

- 4

Originally posted by suyver

Zero is the only solution of the equation [tex] x = -x[/tex].

Not in mod 2 arithmetic.

- #11

turin

Homework Helper

- 2,323

- 3

What is i or π in mod2?

- #12

- 248

- 0

Originally posted by kishtik

That should be

[tex] 0 \; = \; e^{i\pi}+1[/tex]

I hang my head in deep shame. You are (of course) very right.

- #13

matt grime

Science Advisor

Homework Helper

- 9,420

- 4

Originally posted by turin

What is i or π in mod2?

i is defined to be the square root of -1 isn't it? well, then -1=1 (mod 2) and the polynomial

x^2-1 = (x+1)(x+1) mod 2

so the answer is i=1

and n is either 0 or 1 depending on n odd or even resp.

Last edited:

- #14

- 2

- 0

Some people consider 0 to be an asymtote. Not saying i do. but some do.

- #15

- 372

- 0

This can be rendered roughly in English by the following.

I think that is the fundamental question behind this topic.

- #16

turin

Homework Helper

- 2,323

- 3

Not n, π. That is pi.Originally posted by matt grime

... n is either 0 or 1 depending on n odd or even resp.

I'm still trying to understand that bit about i.

- #17

matt grime

Science Advisor

Homework Helper

- 9,420

- 4

if that's all too much, then you probably don't want to know about maximal ideals in the ring of integers

by n I assumed you mean 1+1+1...+1, n times.

The key thing to understand is that when i introduced mod 2 arithmetic, i was pointing out that the question, and many of the answers were assuming that it was posed in the the real numbers. that is notthe only place where zero occurs.

- #18

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 967

Originally posted by quartodeciman[/b]

I like the title of a monograph by nineteenth-century German mathematician Richard Dedekind.

"Was sind und was sollen die Zahlen?"

This can be rendered roughly in English by the following.

What are numbers, and what should they be?"

It sounds a lot better in German, doesn't it. But you are right, it's a cool title.

- #19

turin

Homework Helper

- 2,323

- 3

Matt grime,

It almost seems that you're not listening to me. But first, I will assume that I am very confused about mod2 (being not a math person).

First of all, I thought that mod2 could only deal with two distinct discrete objects, for instance 0 and 1. Then, the combination rule is defined such that 1 + 1 = 0, so that you always stay in the group.

I didn't know that multiplication, exponentiation, imaginary, or irational numbers were allowed in this scheme. I read through your bit about what i is in mod2 over and over again, and I still don't get it. If i = 1, then why write i? Does it only equal 1 when it is considered by itself, but it equals √-1 when it is in a product or exponent? It seems like there is 0 and there is 1, and then there is an infinitude of other "values" (not mod2) that you either have to identify with 0 or with 1. Why not say that i = 0?

I never asked about n. I think you may have seen the html "pi" in my post and interpretted it as an "n." Anyway, I'm curious what you would call pi in mod2. And now I realize that I forgot to ask about e. Both of these are irrational. That seems like it would be a problem for any discrete number system.

Tell me where I have gone way off. You said something about a ring. What is that?

*edit:*

Well, I don't know why I screwed this whole thing up so bad. For some reason, I thought we were talking about Euler's identity. But you were just talking about -x = x this whole time (I just read through the thread again). No wonder you think I'm whacko. I am TERRIBLY sorry about that. I would still like to hear what you have to say about the questions/concerns that I posed above, though.

It almost seems that you're not listening to me. But first, I will assume that I am very confused about mod2 (being not a math person).

First of all, I thought that mod2 could only deal with two distinct discrete objects, for instance 0 and 1. Then, the combination rule is defined such that 1 + 1 = 0, so that you always stay in the group.

I didn't know that multiplication, exponentiation, imaginary, or irational numbers were allowed in this scheme. I read through your bit about what i is in mod2 over and over again, and I still don't get it. If i = 1, then why write i? Does it only equal 1 when it is considered by itself, but it equals √-1 when it is in a product or exponent? It seems like there is 0 and there is 1, and then there is an infinitude of other "values" (not mod2) that you either have to identify with 0 or with 1. Why not say that i = 0?

I never asked about n. I think you may have seen the html "pi" in my post and interpretted it as an "n." Anyway, I'm curious what you would call pi in mod2. And now I realize that I forgot to ask about e. Both of these are irrational. That seems like it would be a problem for any discrete number system.

Tell me where I have gone way off. You said something about a ring. What is that?

Well, I don't know why I screwed this whole thing up so bad. For some reason, I thought we were talking about Euler's identity. But you were just talking about -x = x this whole time (I just read through the thread again). No wonder you think I'm whacko. I am TERRIBLY sorry about that. I would still like to hear what you have to say about the questions/concerns that I posed above, though.

Last edited:

- #20

matt grime

Science Advisor

Homework Helper

- 9,420

- 4

indeed i was only talking about x=-x, nothing to do with e or pi.

Firstly, F_2 is the set {0,1} as you know, with the addition and mult. as you stated. Now, this is all I was using. Now, i is defined as the square root of -1 in C, here there is a analogous element in the field because -1=1 has a square root. Notice however that there is a polynomial

x^2+x+1

that has no root in F_2. Just like in the real/complex case, we simply define A to be a root of this polynomial (the other root is 1+A), then there is an extension F_2[A] which we call F_4, it is the field with 4 elements, it is still not posible to find roots of every polynomial in F_4, so we can extend again and again. Each extension has 2^r elements for some r. The 'limit' of this construction, we'll call F, and it is the algebraic closure of F_2. It is infinite.

In all these fields 1=-1.

In the same way as there is a surjection from Z to F_2, there is a way of relating algebraic integers to "an" algebraic closure of F_2.

Let (2) be the ideal generated by 2 inside the ring of algebraic integers (that is the set of solutions of all monic polynomials with integer coeffs), the ideal is just the set of all things 2x, where x is an algebraic integer. Just like mod 2 arithmetc, we can declare two elments in the alg, integers to be the same if their difference lies in (2) (actually we require a maximal ideal containing (2) but that's a technicality).

Thus it is possible to define the image of any algebraic integer in a field of characteristic 2. However, neither e nor pi are alg. integers so i can't define their images.

To explain the high faluting maximal thing - 2 must get sent to zero, and so sqrt(2) must also get sent to zero if this were to make any sense, but sqrt(2) is not in the ideal (2). The maximal thing there corrects that problem and makes sure that the quotient is a field.

the vast majority of algebraic integers are irrational, in fact the only rational ones are the integers. it's a useful exercise to prove that

interestingly, the golden ratio, a root of x^2+x+1 must get sent to A, or 1+A in this scheme. note that A has no value as a real number! just as i doesn't, it's just a symbol we manipulate according to the rule A^2=A+1

Matt

- #21

turin

Homework Helper

- 2,323

- 3

- #22

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,950

- 19

Here it is in simpler terms.

In the world of complex numbers, a perfectly valid definition of [itex]i[/itex] is:

[itex]i[/itex] is a root of [itex]x^2 + 1[/itex].

All of the arithmetic we do with [itex]i[/itex] can be done algebraically using this fact; for instance:

[tex]

\begin{equation*}\begin{split}

(a + bi) (c + di) &= ac + adi + bic + bidi \\

&= ac + (ad + bc) i + bd (i^2) \\

&= ac + (ad + bc) i + bd (i^2 + 1 - 1) \\

&= ac + (ad + bc) i + bd (0 - 1) \\

&= (ac - bd) + (ad + bc) i

\end{split}\end{equation*}

[/tex]

So, for any practical algebraic purpose, we can simply say that we've simply declared that [itex]i[/itex] is a root of [itex]x^2 + 1[/itex].

(Behind the curtain, there is quite a bit of mathematics involved to prove that such a declaration can be sensible, but you don't need to know what goes on behind the curtain to use such declarations)

Why stop there? We could instead declare [itex]\alpha[/itex] is a root of [itex]x^2 - 2x + 2[/itex]! (1)

This is also a perfectly valid way to create the complex numbers; declare them to be all numbers of the form [itex]p + q \alpha[/itex]. (exercise for those of you at home, and I*strongly* suggest you do it, things will make more sense!: what is [itex](a + b \alpha) (c + d \alpha)[/itex] written in the form [itex]p + q \alpha[/itex]?)

It turns out in this case that (assuming my hasty algebra was correct) if we make the substitution [itex]\alpha \rightarrow i + 1[/itex] that we can convert from this new definition of the complex numbers to the normal definition.

Anyways, we can apply this same idea to any other field, such as the integers mod 2 ([itex]F_2[/itex]). (a field is essentially just a number system in which you can divide by nonzero things)

One could try and define a new number system that is all numbers of the form [itex]p + q \alpha[/itex] where [itex]p, q \in F_2[/itex] and [itex]\alpha[/itex] is a root of [itex]x^2 + 1[/itex]... but that isn't helpful because [itex]x^2 + 1[/itex] already has a root in [itex]F_2[/itex]!!!

But, as before, we can simply pick another polynomial and declare some new field (let's call it [itex]F_2(\alpha)[/itex]) as all numbers of the form [itex]p + q \alpha[/itex] where [itex]p, q \in F_2[/itex] and [itex]\alpha[/itex] is a root of [itex]x^2 + x + 1[/itex]. This system is interesting because [itex]x^2 + x + 1[/itex] doesn't already have a root in [itex]F_2[/itex].

It turns out that when doing this type of construction over a finite field (such as the integers mod a prime), things are*much* more interesting than the boring world of the real and complex numbers.

(that's boring algebraically; they're still very interesting because of their topology)

(1) This sentence was originally incorrectly written as "Why stop there? We could instead declare [itex]\alpha[/itex] is a root of [itex]\alpha^2 - 2\alpha + 2[/itex]!"

In the world of complex numbers, a perfectly valid definition of [itex]i[/itex] is:

[itex]i[/itex] is a root of [itex]x^2 + 1[/itex].

All of the arithmetic we do with [itex]i[/itex] can be done algebraically using this fact; for instance:

[tex]

\begin{equation*}\begin{split}

(a + bi) (c + di) &= ac + adi + bic + bidi \\

&= ac + (ad + bc) i + bd (i^2) \\

&= ac + (ad + bc) i + bd (i^2 + 1 - 1) \\

&= ac + (ad + bc) i + bd (0 - 1) \\

&= (ac - bd) + (ad + bc) i

\end{split}\end{equation*}

[/tex]

So, for any practical algebraic purpose, we can simply say that we've simply declared that [itex]i[/itex] is a root of [itex]x^2 + 1[/itex].

(Behind the curtain, there is quite a bit of mathematics involved to prove that such a declaration can be sensible, but you don't need to know what goes on behind the curtain to use such declarations)

Why stop there? We could instead declare [itex]\alpha[/itex] is a root of [itex]x^2 - 2x + 2[/itex]! (1)

This is also a perfectly valid way to create the complex numbers; declare them to be all numbers of the form [itex]p + q \alpha[/itex]. (exercise for those of you at home, and I

It turns out in this case that (assuming my hasty algebra was correct) if we make the substitution [itex]\alpha \rightarrow i + 1[/itex] that we can convert from this new definition of the complex numbers to the normal definition.

Anyways, we can apply this same idea to any other field, such as the integers mod 2 ([itex]F_2[/itex]). (a field is essentially just a number system in which you can divide by nonzero things)

One could try and define a new number system that is all numbers of the form [itex]p + q \alpha[/itex] where [itex]p, q \in F_2[/itex] and [itex]\alpha[/itex] is a root of [itex]x^2 + 1[/itex]... but that isn't helpful because [itex]x^2 + 1[/itex] already has a root in [itex]F_2[/itex]!!!

But, as before, we can simply pick another polynomial and declare some new field (let's call it [itex]F_2(\alpha)[/itex]) as all numbers of the form [itex]p + q \alpha[/itex] where [itex]p, q \in F_2[/itex] and [itex]\alpha[/itex] is a root of [itex]x^2 + x + 1[/itex]. This system is interesting because [itex]x^2 + x + 1[/itex] doesn't already have a root in [itex]F_2[/itex].

It turns out that when doing this type of construction over a finite field (such as the integers mod a prime), things are

(that's boring algebraically; they're still very interesting because of their topology)

(1) This sentence was originally incorrectly written as "Why stop there? We could instead declare [itex]\alpha[/itex] is a root of [itex]\alpha^2 - 2\alpha + 2[/itex]!"

Last edited:

- #23

matt grime

Science Advisor

Homework Helper

- 9,420

- 4

I'm not sure I can countenance using x to be a root of x^2-2x+2...

- #24

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,950

- 19

While my typo was still technically correct, I will admit it was extraordinarily confusing. Fixed.

- #25

turin

Homework Helper

- 2,323

- 3

I don't see how this could be nonsensible. It is a definition, and I don't see how it could be self-contradictory in this case. I'm assuming we already take for granted what ^2, +, 1, and "root" mean? Then, we just introduce another entry into our dictionary: i. What could be wrong with that?Originally posted by Hurkyl

... for any practical algebraic purpose, we can simply say that we've simply declared that [itex]i[/itex] is a root of [itex]x^2 + 1[/itex].

(Behind the curtain, there is quite a bit of mathematics involved to prove that such a declaration can be sensible, but you don't need to know what goes on behind the curtain to use such declarations)

It doesn't seem to be an issue to continue with replacement. Why stop with the definition for α, when we could instead declare β is a root of xOriginally posted by Hurkyl

Why stop there? We could instead declare [itex]\alpha[/itex] is a root of [itex]x^2 - 2x + 2[/itex]

I don't get it. Isn't it already in that form, that is, two factors of that form?Originally posted by Hurkyl

(exercise for those of you at home, and Istronglysuggest you do it, things will make more sense!: what is [itex](a + b \alpha) (c + d \alpha)[/itex] written in the form [itex]p + q \alpha[/itex]?)

I think I missed what the idea was. You mean we can define things in mod2? I don't get why that is so special. Are you talking about the analogy of extending the real numbers into the complex numbers? Is this FOriginally posted by Hurkyl

Anyways, we can apply this same idea to any other field, such as the integers mod 2 ([itex]F_2[/itex]). (a field is essentially just a number system in which you can divide by nonzero things)

I don't understand this. What does "[itex]p, q \in F_2[/itex]" mean? So what if it's not helpful if it's valid. Does this not prove that i is, by definition, 1 in mod 2? (since the definition of i is the root of xOriginally posted by Hurkyl

One could try and define a new number system that is all numbers of the form [itex]p + q \alpha[/itex] where [itex]p, q \in F_2[/itex] and [itex]\alpha[/itex] is a root of [itex]x^2 + 1[/itex]... but that isn't helpful because [itex]x^2 + 1[/itex] already has a root in [itex]F_2[/itex]!!!

Is this just a random polynomial that you chose as an example that doesn't have roots in mod 2, or is there something special/conventional about it?Originally posted by Hurkyl

But, as before, we can simply pick another polynomial and declare some new field (let's call it [itex]F_2(\alpha)[/itex]) as all numbers of the form [itex]p + q \alpha[/itex] where [itex]p, q \in F_2[/itex] and [itex]\alpha[/itex] is a root of [itex]x^2 + x + 1[/itex]. This system is interesting because [itex]x^2 + x + 1[/itex] doesn't already have a root in [itex]F_2[/itex].

Last edited:

Share: