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

by flyingpig
Tags: daunting, symbol, words
 PF Patron Sci Advisor Thanks Emeritus P: 15,673 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?
P: 2,568
 Quote by micromass 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
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P: 15,673
 Quote by flyingpig 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.
 P: 2,568 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...?
 P: 2,568 Correction * 64
 PF Patron Sci Advisor Thanks Emeritus P: 15,673 Yes, now try to do the other ones. This should be quite easy if you understood this one.
 P: 2,568 oH MY GOD, Please stay with me in case I made a silly mistake. With 64, I will...
P: 2,568
A1 done, A2 speaks for itself.

 Quote by A2 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

 Quote by A3 I sense danger from this one... x + 0 = x Let x = 0 0 + 0 = 0 [b]Let x = 1[/tex] 1 + 0 = 1
 Quote by A4 x + (-x) = 0 Oh boy Let x = 0 0 + (-0) = 0 + 0 = 0 Let x = 1 1 + (-1) =....

DOing M1-M4 D in another post.
 PF Patron Sci Advisor Thanks Emeritus P: 15,673 OK, for A4, you need to use that -1=1 and -0=0. You define this to be so.
P: 2,568
 Quote by M1 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
 Quote by M2 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
 Quote by M3 Other than 0, there is a 1 such that x * 1 = x Let x = 0 0 * 1 = 0 Let x = 1 1*1 = 1
 Quote by M4 Other than 0, we have an inverse for x Oh wait...should I just do x = 1 case...?
 Quote by D NOOO ANOTHER 8 CASE. I'll be back on this one...
Need water...be back in 2 mins.
 PF Patron Sci Advisor Thanks Emeritus P: 15,673 The second of M1 is incorrect.
P: 2,568
Yes because 0*0 is not 1...!

 Quote by D 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
 PF Patron Sci Advisor Thanks Emeritus P: 15,673 Looks ok!!
 P: 2,568 Now just imagine I have to put all of this on paper.
P: 2,568
 Quote by SammyS 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.
 Quote by micromass 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?
 PF Patron Sci Advisor Thanks Emeritus P: 15,673 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?
 P: 2,568 Oh is it from 1 + 1 = 0, that 1 = -1? oh okay
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