
#37
Sep2611, 08:01 PM

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P: 16,543

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? 



#38
Sep2611, 08:04 PM

P: 2,568

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 



#39
Sep2611, 08:12 PM

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P: 16,543





#40
Sep2611, 08:14 PM

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...? 



#41
Sep2611, 08:17 PM

P: 2,568

Correction * 64




#42
Sep2611, 08:19 PM

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P: 16,543

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




#43
Sep2611, 08:20 PM

P: 2,568

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




#44
Sep2611, 08:25 PM

P: 2,568

A1 done, A2 speaks for itself.
DOing M1M4 D in another post. 



#45
Sep2611, 08:33 PM

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P: 16,543

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




#46
Sep2611, 08:40 PM

P: 2,568





#48
Sep2611, 09:01 PM

P: 2,568

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




#50
Sep2611, 09:09 PM

P: 2,568

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




#51
Sep2611, 09:53 PM

P: 2,568





#52
Sep2611, 09:55 PM

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P: 16,543

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? 



#53
Sep2611, 09:58 PM

P: 2,568

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



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