Originally posted by eatmybishopSince when is "+" used as a symbol for OR operations? You also said that lausey's boolean algebra was incorrect, which is wasn't. But yes, you are right, if you use OR instead of AND, your table is correct.
if its based on or operations its right... please tell me where that is wrong
Originally posted by Nordlysthe + is used as a representation of the or operations; just like 'and' is represented by . and 'not' is '
Since when is "+" used as a symbol for OR operations? You also said that lausey's boolean algebra was incorrect, which is wasn't. But yes, you are right, if you use OR instead of AND, your table is correct.
i.e...
AND Operations ·
0·0 = 0 A·0 = 0
NOT Operations '
0' = 1 A'' = A
a little out of depth here nordlys... maybe you need to join the others
Originally posted by eatmybishopI am used to the notation where V is used for OR, an upside-down V for AND, and something looking approximately like a forged -, for NOT. But I looked it up (probably have actually seen it before, so should have remembered), and your use of + is also correct - good for you, at least you got one thing right! I can't find a source for your way of representing "NOT", but I won't exclude the possibility that some people use that, too.
the + is used as a representation of the or operations; just like 'and' is represented by . and 'not' is '
i.e...
AND Operations ·
0·0 = 0 A·0 = 0
NOT Operations '
0' = 1 A'' = A
a little out of depth here nordlys... maybe you need to join the others
Lausey's table is still correct.
Originally posted by Nordlyswow! the first (and no doubt the last) person to say i got something right on rhp.... doesnt happen often so i'll make the most of it...
I am used to the notation where V is used for OR, an upside-down V for AND, and something looking approximately like a forged -, for NOT. But I looked it up (probably have actually seen it before, so should have remembered), and your use of + is also correct - good for you, at least you got one thing right! I can't find a source for your way of representing xclude the possibility that some people use that, too.
Lausey's table is still correct.
is he/she right..?
he/she said 1 + 1 = 0 but the + represents the or operation, therefore, it doesnt, 1 + 1 = 1 if boolean algebra logical expressions are used in their natural form
Originally posted by Palynkasorry, but i think you have no idea what you're saying, you cannot define + to 'and' in boolean algebra... that's like saying i define = to + in decimal notation
Because lausey defined + to be the operation 'AND'. Therefore, the algebra contained in his post is correct.
Edit - Actually, he wrote AND, so notation isn't even a question.
Originally posted by Palynkai take it back, you are right there, he didnt say that, my mistake; but he did say 1 + 0 = 0 where is should read 1 + 0 = 1... that is because the + represents the or operation in boolean algebra... he/she should have written 1·0 = 0 as it's the - that represents the and operation...
Wrong.
you keep on saying i'm wrong yet don't give an explanation; could you please tell me where i'm wrong on this and i'll happily back down
Originally posted by eatmybishopYou want to try programming in C++. You can define + to be whatever the hell you want it to be. And = for that matter 🙂
sorry, but i think you have no idea what you're saying, you cannot define + to 'and' in boolean algebra... that's like saying i define = to + in decimal notation
Originally posted by eatmybishop?
sorry, but i think you have no idea what you're saying, you cannot define + to 'and' in boolean algebra... that's like saying i define = to + in decimal notation
OK. I define '=' as add
I define '*' as equals
1=1*2
Its no big deal. Simple operations like addition have standardised symbols but other mathematical concepts have a variety of ways of writing.
Consider all the ways of writing differentials.
(y' = dy/dt et cetera)
Mathematicians define things all the time (axioms) in order to get more interesting stuff out (theorems).
Who defined '=' as equals to start with? God?
Originally posted by eatmybishopHe said:
i take it back, you are right there, he didnt say that, my mistake; but he did say 1 + 0 = 0 where is should read 1 + 0 = 1... that is because the + represents the or operation in boolean algebra... he/she should have written 1·0 = 0 as it's the - that represents the and operation...
you keep on saying i'm wrong yet don't give an explanation; could you please tell me where i'm wrong on this and i'll happily back down
"1 AND 0 = 0"
He did not say:
"1 + 0 = 0"
Look at his post.
Edit - Well said, wolfgang.
Originally posted by mtthwyes you can, just like you can if you program in c or java... do you honestly think the cpu architecture sees it that way; of course not, i also program in assembly; c++, java, c etc will have their statements converted to an in-memory layout via the compiler using the natural expressions of boolean algebra... did you learn anything about c++???
You want to try programming in C++. You can define + to be whatever the hell you want it to be. And = for that matter 🙂
Originally posted by wolfgang59but you cant do that with a cpu architecture, the cpu wasnt designed that way; to us we think we can but the cpu architecture will always convert it back to its natural state
?
OK. I define '=' as add
I define '*' as equals
1=1*2
Its no big deal. Simple operations like addition have standardised symbols but other mathematical concepts have a variety of ways of writing.
Consider all the ways of writing differentials.
(y' = dy/dt et cetera)
Mathematicians define things all the time (axioms) in order to get more interesting stuff out (theorems).
Who defined '=' as equals to start with? God?