Not only is the symbol daunting, but the words are too

[flyingpig]

Homework Statement

Check that \mathbb{F}_2 is a field

The Attempt at a Solution

The problem set given to us is mixed in notes, so I might have missed something because it's so messy.

These are the two properties given.

A field F is a set with + and * on it such that

(x,y) \to x + y \in \mathbb{F} + : \mathbb{F} \times \mathbb{F} \to \mathbb{F}

(x,y) \to xy \in \mathbb{F} \cdot : \mathbb{F} \times \mathbb{F} \to \mathbb{F}

The notes previous saidly something about "a curious field F = {0,1}"

What do they mean by "curious field"?

 

[HallsofIvy]

It means, believe it or not, a field that is, in some way, "curious" (in the sense of "odd", "unusual" or "unique"). What's curious about that field is that it is the smallest possible field.

 

[flyingpig]

It means, believe it or not, a field that is, in some way, "curious" (in the sense of "odd", "unusual" or "unique"). What's curious about that field is that it is the smallest possible field.

I thought {0,0} is smaller...?

How do I manipulate the properties of the fields to help me with the question?

 

[flyingpig]

My prof said to verify A1 to A4

[PLAIN]http://img194.imageshack.us/img194/2548/unledtao.png

[PLAIN]http://img717.imageshack.us/img717/2285/unledxig.png

and M1 to M4 and D (but not prop 2.7)

Does that sound right? How does that relate to the "curious field"? I thought those axioms hold for all fields

 

[SammyS]

I thought {0,0} is smaller...?

By the set {0, 0} you must mean the set {0}

This can't be used to form a field. It can't even be used to form a ring. There is no multiplicative identity element in {0} \ {0}, because {0} \ {0} is the empty set. (See M3.)

 

[SammyS]

My prof said to verify A1 to A4
...

and M1 to M4 and D (but not prop 2.7)

Does that sound right? How does that relate to the "curious field"? I thought those axioms hold for all fields

Of course they apply to all fields.

Do you have addition and multiplication tables for the "curious field" ?

 

[flyingpig]

I am digging, one sec.

 

[flyingpig]

[PLAIN]http://img707.imageshack.us/img707/652/unledfz.jpg

We were given this. Most of them seem okay except 1 + 1 = 0...

 

[SammyS]

What's not OK with 1 + 1 = 0 ?

 

[flyingpig]

Doesn't A2 say 1 + 1 = 1 + 1 and not 0...?

 

[SammyS]

If 1+1 = 1, then 1 has no additive inverse.

 

[flyingpig]

If 1+1 = 1, then 1 has no additive inverse.

No big words please...

 

[SammyS]

If 1 + 1 = 1, then there is no inverse for 1 using the addition operation. In other words, there would be no way to fulfill A4 .

 

[flyingpig]

Let's slow down a bit and I need something clarified. Through 2.45 to 2.52, do I have to verify all of them to show that F_2 is indeed a field?

I just started field so I am still a novice.

When they say F_2 = {0,1}, they really want us to show every combination of addition and multiplication of the numbers 0 and 1?

I still understand how that makes 1 + 1 = 1, it shuold be really, not 1 or 0.

What does having an inverse have to do with having a unique -x? (x being 1)

 

[SammyS]

What does having an inverse have to do with having a unique -x? (x being 1)

For this field, -1 = 1 . This is because the thing you need to add to 1 to give you the identity element for addition is 1.

 

[flyingpig]

For this field, -1 = 1 . This is because the thing you need to add to 1 to give you the identity element for addition is 1.

This word "field" is being abused way too much to bypass basic addition rules...

Let's just pick on that guy and the first one, 0 + 0 = 0 first (I am guessing := can
be replaced by = here?)

All I have to do is verify right?

So by A2 0 + 0 = 0 + 0 = 0

which is true (order does not matter)

I can't get my way around 1 + 1 = 1 still because I still thinking this in basic addition that 1 + 1 = 2 and you are telling me that this "field" has such a property that -1 = 1, a lot of conflict in my mind.

 

[SammyS]

There is no such element as 2 in the set {0, 1}.

This field, F₂ a.k.a. the "curious field", is the set {0,1} with the operations, + and ∙ , defined in post #8. The ":=" combination is used to indicate a definition. Thus, 2.45 through 2.52 are definitions for the field, F₂.

For this field, F₂, you cannot have 1 + 1 = 2. If you did have 1 + 1 = 2, then F₂ would not be closed for the + operation, would not be a field.

Some important WORDS (Not all are big, but the ideas have big importance.):

set
binary operation
closed , closure
identity -- as in identity element, identity function, identity matrix ...

additive identity --> the identity element for the addition operation. Your text denotes this as the numeral, 0 .
multiplicative identity --> the identity element for the multiplication operation. Your text denotes this as the numeral, 1 .​

group, ring, field
negative -- otherwise known as the opposite or the additive inverse
inverse -- your text uses this for the multiplicative inverse which is often called the reciprocal.

The word inverse is also used to describe functions, matrices, binary operations, etc.​

associative
commutative
distributive
...​

 

[HallsofIvy]

I thought {0,0} is smaller...?

How do I manipulate the properties of the fields to help me with the question?

What do you mean by "{0, 0}"? {0}? Most definitions of "field" require that there be at least two members.

 

[flyingpig]

There is no such element as 2 in the set {0, 1}.

This field, F₂ a.k.a. the "curious field", is the set {0,1} with the operations, + and ∙ , defined in post #8. The ":=" combination is used to indicate a definition. Thus, 2.45 through 2.52 are definitions for the field, F₂.

For this field, F₂, you cannot have 1 + 1 = 2. If you did have 1 + 1 = 2, then F₂ would not be closed for the + operation, would not be a field.

Some important WORDS (Not all are big, but the ideas have big importance.):

set
binary operation
closed , closure
identity -- as in identity element, identity function, identity matrix ...

additive identity --> the identity element for the addition operation. Your text denotes this as the numeral, 0 .
multiplicative identity --> the identity element for the multiplication operation. Your text denotes this as the numeral, 1 .​

group, ring, field
negative -- otherwise known as the opposite or the additive inverse
inverse -- your text uses this for the multiplicative inverse which is often called the reciprocal.

The word inverse is also used to describe functions, matrices, binary operations, etc.​

associative
commutative
distributive
...​

Oh okay, so to verify (the original problem of this thread) all of those properties, i must use A1 - A4 and M1 - M4, D

But how do I do that properly? PLease help!

Like for the first one, you have 0 + 0 := 0, that doesn't fall into any of the axioms.

The one that even looks similar is A2, i.e.

0 + 0 = 0 + 0 = 0

 

[micromass]

Oh okay, so to verify (the original problem of this thread) all of those properties, i must use A1 - A4 and M1 - M4, D

No, you mustn't use A1-A4, M1-M4 and D. You must verify them.
For example, for A1, you must check that

x+(y+z)=(x+y)+z

So, take 3 arbitrary elements of \mathbb{F}_2 and go check associativity.

 

[flyingpig]

Three elements? There are only two in {0,1}

 

[Mark44]

x, y, and z can each take on one value from the set {0, 1}.

 

[micromass]

Three elements? There are only two in {0,1}

Just take three elements (which can coincide)

So for example, you can take x=0,y=0,z=0. Or x=1, y=0, z=0. Etc.
So for every choice of three elements, check that

x+(y+z)=(x+y)+z

 

[flyingpig]

So for instance.

A1

x + (y + z) = (x + y) + z

Oh okay.

A1

x + (y + z) = (x + y) + z

Let x = 0, y = 0, z = 0

0 + (0 + 0) = (0 + 0) + 0 = 0 + 0 = 0

Thus A1 is true

 

[micromass]

So for instance.

A1

x + (y + z) = (x + y) + z

Oh okay.

You must show it for all x,y,z. You cna't just show it for x=y=z=0. You must show it for all 9 choices of x,y and z.

 

[flyingpig]

You must show it for all x,y,z. You cna't just show it for x=y=z=0. You must show it for all 9 choices of x,y and z.

So this can't work

x = 0, y = 0, z = 0
x = 0, y = 1, z = 1
x = 0, y = 1, z = 0

Does x have to be 1 at least once?

 

[micromass]

So this can't work

x = 0, y = 0, z = 0
x = 0, y = 1, z = 1
x = 0, y = 1, z = 0

Does x have to be 1 at least once?

Yes, you must let x,y,z run through all possible values. There are 9 combinations in total.

 

[flyingpig]

x + (y + z) = (x + y) + z

Let x = 0, y = 0, z = 0

0 + (0 + 0) = (0 + 0) + 0 = 0 + 0 = 0

Let x = 0, y = 0, z = 1

0 + (0 + 1) = (0 + 0) + 1 = 0 + 1 = 1

Let x = 0, y = 1, z = 0

0 + (1 + 0) = (0 + 1) + 0 = 1 + 0 = 1

Let x = 1, y = 0, z = 0

1 + (0 + 0) = (1 + 0) + 0 = 1 + 0 = 1

Let x = 1, y = 1, z = 0

1 + (1 + 0) = (1 + 1) + 0 = 1 + 0 = 1 <=== doubts...

Let x = 1, y = 0, z = 1

1 + (0 + 1) = (1 + 0) + 1 = 1 + 1 = 0 <=== doubts...

Let x = 0, y = 1, z = 1

0 + (1 + 1) = (0 + 1) + 1 = 1 + 1 = 0 <=== doubts...

Let x = 1, y = 1, z = 1

1 + (1 + 1) = (1 + 1) + 1 = 0 + 1 = 1

I seem to be missing a combination...

 

[micromass]

Let x = 1, y = 0, z = 1

1 + (0 + 1) = (1 + 0) + 1 = 1 + 1 = 0 <=== doubts...

This is wrong.

I seem to be missing a combination...

You are correct. There are only 8 combinations. A mistake of mine.

 

[flyingpig]

This is wrong.

2.46 and 2.47? I had my doubts...

 

[micromass]

2.46 and 2.47? I had my doubts...

What in Earth do you mean with 2.46 and 2.47?

 

[flyingpig]

What in Earth do you mean with 2.46 and 2.47?

[PLAIN]http://img707.imageshack.us/img707/652/unledfz.jpg

Sorry lol, I was reading in my mind.

 

[micromass]

And how would you evaluate

1 + (0 + 1)

using those rules?

 

[flyingpig]

1 + (0 + 1) = 1 + 1 = 0

 

[micromass]

And how would you evaluate (1+0)+1 ?? Is it also 1??

 

[flyingpig]

And how would you evaluate (1+0)+1 ?? Is it also 1??

(1 + 0) = 1 from 2.46

so I get 1 + 1 = 0 still

Also RedBelly I was actualyl going to post my answers lol and then it said "Sorry! the thread is locked!"

Never mind d) is 375. They all move up by 75...each. Stupid question.

This was solutiuon they gave me

[PLAIN]http://img849.imageshack.us/img849/4906/unledhsb.png

Which just means the answer is a multiple of 5, I thought they meant the contours are "5M units apart"

 

[micromass]

OK, so that is indeed correct. We indeed have that

1 + (0 + 1) = (1 + 0) + 1 = 0

But that is not were the error was. The error is in the following

1 + (1 + 0) = (1 + 1) + 0 = 1 + 0 = 1

So can you calculate

1+(1+0)=...
(1+1)+0=...

again and see where the error is?

 

[flyingpig]

So can you calculate

1+(1+0)=...
(1+1)+0=...

again and see where the error is?

But they all arrive at the same answer

1 + (1 + 0) = 1 + 1 = 0

(1 + 1) + 0 = 0 + 0 = 0

But 1 + (1 + 0) from A2 says 1 + (1 + 0) = (1 + 1) + 0 = 0 + 0

I don't see the error

 

[micromass]

But they all arrive at the same answer

1 + (1 + 0) = 1 + 1 = 0

(1 + 1) + 0 = 0 + 0 = 0

But 1 + (1 + 0) from A2 says 1 + (1 + 0) = (1 + 1) + 0 = 0 + 0

I don't see the error

Indeed, they are all 0. But before you said that it was equal to 1. So now you have the correct answer.

 

[flyingpig]

Impossible I had 0! It was in post#29!

Oh well at least I got the other 7 right! ahahhaah

Wait, does that mean I will have 8 x 8 = 64 lines to write for all the other ones...?

 

[flyingpig]

Correction * 64

 

[micromass]

Yes, now try to do the other ones. This should be quite easy if you understood this one.

 

[flyingpig]

oH MY GOD, Please stay with me in case I made a silly mistake. With 64, I will...

 

[flyingpig]

A1 done, A2 speaks for itself.

x + y = y + x

Let x = 0, y = 0

0 + 0 = 0 + 0 = 0

Let x = 0, y = 1

0 + 1 = 1 + 0 = 1

Let x = 1, y = 0

1 + 0 = 0 + 1 = 1

Let x = 1, y = 1

1 + 1 = 1 + 1 = 0

OKay probably not 64 then

I sense danger from this one...

x + 0 = x

Let x = 0

0 + 0 = 0

Let x = 1[/tex]

1 + 0 = 1

x + (-x) = 0

Oh boy

Let x = 0

0 + (-0) = 0 + 0 = 0

Let x = 1

1 + (-1) =...

Stuck already...I am an idiot

DOing M1-M4 D in another post.

 

[micromass]

OK, for A4, you need to use that -1=1 and -0=0. You define this to be so.

 

[flyingpig]

Damn it, 8 cases again

(xy)z = x(yz)

Let x = 0, y = 0, z = 0

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

Let x = 0, y = 0, z = 1

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

Let x = 0, y = 1, z = 0

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

Let x = 1, y = 0, z = 0

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

Let x = 1, y = 1, z = 0

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

Let x = 1, y = 0, z = 1

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

Let x = 0, y =1, z = 1

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


Let x = 1, y = 1, z = 1

(1*1)1 = 1(1*1) = 1*1 = 1

xy = yx

Let x = 0, y = 0

0*0 = 0*0 = 0

Let x = 0, y = 1

0*1 = 1*0 = 0

Let x = 1, y = 0

1*0 = 0*1 = 0

Let x = 1, y = 1

1*1 = 1*1 = 1

Other than 0, there is a 1 such that x * 1 = x

Let x = 0

0 * 1 = 0

Let x = 1

1*1 = 1

Other than 0, we have an inverse for x

Oh wait...should I just do x = 1 case...?

NOOO ANOTHER 8 CASE. I'll be back on this one...

Need water...be back in 2 mins.

 

[micromass]

The second of M1 is incorrect.

 

[flyingpig]

Yes because 0*0 is not 1...!

Let x = 0, y = 0, z = 0

0(0 + 0) = 0*0 + 0*0 = 0 + 0 = 0

Let x = 0, y = 0, z = 1

0(0 + 1) = 0*0 + 0*1 = 0 + 0 = 0

Let x = 0, y =1, z = 0

0(1 + 0) = 0*1 + 0*0 = 0 + 0 = 0

Let x = 1, y = 0, z = 0

1(0 + 0) = 1*0 + 1*0 = 0 + 0 = 0

Let x = 1, y = 1, z = 0

1(1 + 0) 1*1 + 1*0 = 1 + 0 = 1

Let x = 1, y = 0, z = 1

1(0 + 1) = 1*0 + 1*1 = 0 + 1 = 1

Let x = 0, y = 1, z = 1

0(1 + 1) = 0*1 + 0*1 = 0 + 0 = 0

Let x = 1, y = 1, z = 1

1(1 + 1) = 1*1 + 1*1 = 1 + 1 = 0

I am pretty confident in this one

 

[micromass]

Looks ok!

 

[flyingpig]

Now just imagine I have to put all of this on paper.