Results 16 to 27 of 27

- June 21st 2014, 08:45 AM #16

- Joined
- Nov 2010
- Posts
- 1,961
- Thanks
- 798

- June 21st 2014, 10:23 AM #17

- Joined
- Nov 2013
- From
- California
- Posts
- 2,789
- Thanks
- 1149

- June 21st 2014, 03:05 PM #18

- Joined
- Mar 2011
- From
- Tejas
- Posts
- 3,401
- Thanks
- 763

## Re: 1+1=0?

The original post lists the axioms for a field. Not very well, but I assume they are being copied from a book somewhere, where the language is probably better. In fields, the existence of $-a$ for every $a$, and $\dfrac{1}{a}$ for every $a \neq 0$ is "part of the deal".

It is NOT a theorem in ALL fields that $1 + 1 = 0$ and similarly for $1 + 1 \neq 0$. In some fields the first is true, and the second is false (these fields are called characteristic 2 fields), and in some fields the first is false, and the second is true (like the rational numbers, for example). The characteristic of a field is not part and parcel of the definition of a field: different fields may have different characteristics (analogous to how different vector spaces may have different dimensions).

I sincerely doubt this forum as an organization is "attacking" your computer. I do suspect you may have picked up some malware somewhere, and I recommend doing an anti-virus full-disk scan after booting up into safe mode. I tend to carpet-bomb with Kaspersky, AVG, Malwarebytes, Spybot S&D, etc. and even then some stuff gets through.

vBulletin software (which I assume this forum uses) CAN recognize IP addresses (unless you are using an anonymizer proxy, it's not that difficult for your IP to be discovered, after all the server at an internet IP address needs to know where to send the web pages TO), but most ISP's assign "public" IP addresses dynamically, which means you get a new one every time you re-connect to the internet. My guess is that the people who maintain the library computers aren't that savvy about cookie management, and do not regularly flush the internet cache of the machines there. Just a guess, but if that is so, it could be that ANYONE who used the same library computer you did, and navigated to this site, might automatically be logged on as you.

You should be able to disable this in your user control panel, but you will have to enter your username and password for EVERY visit. If you notice any evidence of someone tampering with your account, try PMing Dan (topsquark).

- June 22nd 2014, 10:48 AM #19
## Re: 1+1=0?

Yes, I can identify IPs. But as I have told you several times we at the Forum have not sent you a virus, nor have we been blocking your access to certain threads. It's not all about you, others are having problems as well.

Since you've apparently been ignoring my (multiple) PMs on this issue, just how do you want me to handle this? I'm trying to contact mash, who apparently gone AWOL again, as I've told you. Stop being paranoid.

Why don't you think it over for a week.

-Dan

- June 30th 2014, 10:40 AM #20

- Joined
- Aug 2010
- Posts
- 961
- Thanks
- 98

## Re: 1+1=0?

The answer to OP is yes, 1+1 = 0 is possible, depending on the field definitions.

For ex, define a,b=0,1 and 1+1=0. Then by suitable definition, the other axioms are satisfied.

For ex, how would you define 1+0 to satisfy the Field Axioms?

The key concept is definition. Once you accept it, there is nothing unnatural about 1+1=0. It is unnatural only from the perspective of what we are familiar with.

Abstract examples or terminology without definitions, or reference to them, is meaningless and annoying at the level of this OP.

Some other examples of Fields:

Ex1: 4-element field: Field (mathematics) - Wikipedia, the free encyclopedia.

Note 1+1=0 in this field also.

Ex2: Zp: a,b,c.. are 0,1,2…p-1. a+b = a+b modp, axb =axb modp, (g modp is remainder after division of g by p) and it can be verified the field axioms are satisfied.

Ex3: a+b√2, a,b rational

a+b√2=c+d√2 if a=c and b=d

0=0+0√2

1=1+0√2

(a+b√2)+(c+d√2)=(a+c)+(b+d)√2

(a+b√2)(c+d√2)=[(ac+2bd)+(b+d)√2]

(a+b√2)X=c+d√2, multiply both sides by a-b√2 to solve for X.

Romsek, thanks for your solicitation. Got help for my paranoia from Verizon Tech Support in India. The guy was great: not only got me back on the internet (got an answer), but insisted on looking for the root cause and found it, the true scientific attitude. I am relying on what I have read of “Malwarebytes” to completely cure my paranoia. Hope it works if needed.

- June 30th 2014, 11:17 AM #21

- Joined
- Aug 2010
- Posts
- 961
- Thanks
- 98

## Re: 1+1=0?

Ref post # 9

F(0,1,a,a+1) Is meaningless.

Ie, given F(a,b,c,d) you can’t define d by the field. d is part of the definition of the field. It’s used to define the field. The elements used to define a field can’t be defined by the field. It’s a circular definition.

- June 30th 2014, 12:29 PM #22

- Joined
- Nov 2010
- Posts
- 1,961
- Thanks
- 798

## Re: 1+1=0?

I agree, is meaningless. However, that was not written anywhere in post #9. Deveno wrote . What you wrote appears to be a function that takes four variables as input. Deveno wrote is a set of four elements. He then explained the addition and multiplication for the elements of that set (in other words, he defined binary operators and on ). He then showed how you can check that the set along with its two binary operators forms a field (by verifying the field axioms). This is a standard construction for the field of four elements. I saw similar constructions in both a text by Dummit and Foote as well as a text by Rotman. If you want me to look it up, I can probably give you exact page numbers for you to see it in print if you prefer.

- July 1st 2014, 04:34 AM #23

- Joined
- Aug 2010
- Posts
- 961
- Thanks
- 98

## Re: 1+1=0?

By F(0,1,a,a+1) I obviously meant the field F={0,1,a,a+1} which says:

a+1 is the sum of a and 1.

but what is the sum of a and 1? a+1

the classic circular definition.

F={0,1,a,b} allows me to define a and b as I like.

F={0,1,a,a+1} only allows me to define a. What is a+1?

Ie, F={0,1,a,a+1} is undefined and therefore meaningless, as is everything else that follows in the subject post.

- July 1st 2014, 04:48 AM #24

- Joined
- Nov 2010
- Posts
- 1,961
- Thanks
- 798

## Re: 1+1=0?

Read up on free groups and a+1 makes sense. Again, the notation used is a standard notation. Arguing on this forum that standard notation is undefined is rather... fruitless? I mean, it is not an argument with us. It is an argument with a much larger mathematical community that has accepted the notation as standard.

- July 1st 2014, 06:42 AM #25

- Joined
- Mar 2011
- From
- Tejas
- Posts
- 3,401
- Thanks
- 763

## Re: 1+1=0?

This is a bit of a detour, but I hope it does someone some good.

Most young children are familiar with the expression:

1 + 1 = 2 (Not using any "fancy exotic" system, here. These are natural numbers).

If you ask many mathematicians, they might say: "that's not an equation, it's a definition".

One of the notable features about the natural numbers is its RECURSIVE nature: we define "bigger numbers" in terms of "smaller numbers we already know". The main tool we have in this regard is called the SUCCESSOR function, which, given a natural number $k$ ($k$ is just a symbol here) gives us "the next natural number" (it emulates formally the process of "counting" as we might do on our fingers, or using stones).

Now addition is also defined "recursively", by:

$k + 0 = k$

$k + s(m) = s(k + m)$, where $s$ denote the successor function.

If $m$ is the natural number 0 (the only natural number we "start out with"), this gives us:

$k + s(0) = s(k + 0) = s(k)$.

The SYMBOL 1 is assigned to $s(0)$, by long-standing historical convention (thank the ancient Indians and Arabs for this, and also that Fibonacci's father was a well-traveled merchant, who picked up lots of odd trivia from his trade).

So we may re-write the above as:

$k + 1 = s(k)$.

This means we can use "k + 1" as a stand-in for s(k). In particular, letting $k = s(0) = 1$, we have:

$1 + 1 = s(1) = 2$ <---another historical convention.

If we wanted, we could keep using "1 + 1" instead of the symbol 2, although this would take up much more space, when performed arithmetic.

When creating an "addition table" for the natural numbers, there would be nothing wrong with using 1+1 instead of 2 in the table, but it doesn't tell us "which other natural number" 1+1 is.

But I see you are not satisfied that defining the 4-element field as I have isn't "circular" (it's actually tautological, which is slightly different). We could take a different approach, perhaps this will be more amenable to you:

Define $F$ as the vector space over the field $\Bbb Z_2$ (which has only two elements, 1 and 0) of dimension 2. This gives us 4 elements:

$F = \{(0,0),(1,0),(0,1),(1,1)\}$.

The addition should be clear: we are taking (1,0) and (0,1) as a basis, and using the vector space addition. What is not clear at the outset, is that we can define a multiplication on this set as well:

(0,0)*(0,0) = (0,0)

(0,0)*(1,0) = (0,0)

(0,0)*(0,1) = (0,0)

(0,0)*(1,1) = (0,0)

(1,0)*(1,0) = (1,0)

(1,0)*(0,1) = (0,1)

(1,0)*(1,1) = (1,1)

(0,1)*(0,1) = (1,1)

(0,1)*(1,1) = (1,0)

(1,1)*(1,1) = (0,1) and the products not listed use commutativity.

Again, it is tedious to show that this multiplication is associative and that it distributes over the vector addition, but it can be done.

If we use the $\Bbb Z_2$-linear mapping $L: F \to \{0,1,a,a+1\}$ (as in my previous posts) given by:

$L((1,0)) = 1$

$L((0,1)) = a$, we obtain an isomorphism of vector spaces, and you can verify for all 16 possible products that $L(v_1\ast v_2) = L(v_1)\ast L(v_2)$, so that these are equivalent field structures (only the names have been changed, to protect the innocent).

There is a 3rd construction of $F$ as well, that uses a quotient ring of $\Bbb Z_2[x]$ modded by the ideal generated by the polynomial $x^2 + x + 1$, but I think you might not like all the work you have to do to make it "pay off".

- July 2nd 2014, 12:58 PM #26

- Joined
- Aug 2010
- Posts
- 961
- Thanks
- 98

## Re: 1+1=0?

The two pluses in the “Field” F = {0,1,a,1+a, +,x} obviously denote the same operation ( same symbol), in which case F is not a field (see above) and post #9 is meaningless.

If the first plus means something else, it should be denoted x_{1}and a+_{1}1 defined. There is the trivial definition that it is just a symbol for a fourth element which can be arbitrarily assigned. In any case it doesn’t apply to post #9, which doesn’t differentiate symbols.

I note, following a suggestion of SlipEternal’s, that Dummit and Foote define F(a,b,c,d) as the field generated by a,b,c,d. 2nd Ed, Amazon “Look Inside” and there is a symbol table. (Granted, I am making an assumption about what “generated” means in this case (generated not in index)). Almost wavered on buying a used version of 2nd ed but 1000pgs is ridiculous for a bloated B&M, judging from Table of Contents. That’s why I like older books, and they're better made.

Back to the public computer. Long walk on a hot day.

- July 2nd 2014, 01:18 PM #27