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

[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.

 

[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.

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

I overlooked something. since WHEN DID I DEFINE -1 = 1?

 

[micromass]

Well, you need to find for each x, an element y such that

x+y=0

By definition, this element is -x.

So, for x=0, you need an element such that 0+y=0. When you look at the axioms, you see that y=0 is the only possibility. So -0=0.

Now, what is -1?

 

[flyingpig]

Oh is it from 1 + 1 = 0, that 1 = -1? oh okay

 

[micromass]

Oh is it from 1 + 1 = 0, that 1 = -1? oh okay

Yes.

 

[flyingpig]

I am still stuck on M4...

 

[micromass]

Take x in your field. How would you define $x^{-1}$??

 

[flyingpig]

$$xx^{-1} = 1$$

$$x^{-1} = 1/x$$

That doesn't change the original definition...

I still can't use 0.

 

[flyingpig]

Oh wait, M4 actually states for all x in R, without 0, the axiom holds, I don't even need to test x = 0 right?

Ah shoot for M3, I put 0 * 1 = 0, that's wrong isn't it? Because it says x without 0

 

[micromass]

Sigh...

For each x nonzero, you need to find a y such that xy=1.
If x=1. What can you take as y?

 

[micromass]

Oh wait, M4 actually states for all x in R, without 0, the axiom holds, I don't even need to test x = 0 right?

Right.

 

[flyingpig]

This is probably another stupid question.

Since M3 states the same condition, it would be wrong/unnecessary to even do 0*1 = 0 because it contradicts M4 \ {0} right?

 

[micromass]

M3 does not state the same thing as M4...

 

[flyingpig]

M3 does not state the same thing as M4...

But they have a similar condition...

 

[flyingpig]

Oh wait you know what, never mind. I was looking at 2.49 and I thought $$0^{-1} = 0$$ and I had to test $$0*1 = 0$$.

So yes I do need to test 0*1 = 0 for x =0