Hi, I have this proof I have to do, and I just don't know how to star proofs off.
Here is the question:
Show that the operation of subtraction is neither associative nor commutative.
If anyone can help with this, I would appreciate it.
David.
Hi, I have this proof I have to do, and I just don't know how to star proofs off.
Here is the question:
Show that the operation of subtraction is neither associative nor commutative.
If anyone can help with this, I would appreciate it.
David.
Is that it? Isn't that just stating what we have to show though?
I thought there would be a little more work than that? Or is that what you have given me to start off. If that's the case, I already had that bit.
I got that from the question.
Any more to this?
here is the thing with these definition in the textbook. they are only true for positive integers. by the definition and introduction of negative quantities in algebra these rules in the textbooks are just shabby nonsense. once they show you the rules. they go about breaking every rule.
we have the numeric constant the variable the sign and the operation. variables have undetermined sign. so we leave out sign in the examples. every operand has an operation associated with it. the leading operands have an operation which is normally left off but I display it here.
alexmahone's expressions are not equivalent unless you say there is no negative integers in you algebra. that is quite a large limitation. and as far as I am aware they do allow the use of negative numbers in all the books that state these rules. so to use associativity and commutativity to transform your expressions you would have to do this.
and by commutativity you would do this.
now they have mixed all of this up and give you some 'extra' rules about removing associative braces preceded by a subtraction operation where you magically change sign! subtraction signs magically travel with there operand! and you have 'extra' rule about negative fractions where you magically change sign! but is not a fraction just division of operands grouped with associativity?
you can sort this out if you try to simplify expressions and every step list by what law you do that step. you will soon find out what you can and can not do.
To prove a general statement is NOT true, it is sufficient to give a single "counter example"- one example to show that the statement is not true for that example and so not "always" true.
That is what alexmahone and Plato are suggesting:
For subtraction to be commutative, it would have to be true that "a- b= b- a" for all numbers, a , b, c. But (5- 4)= 1 and (4- 5)= -1 shows that is NOT true for those numbers and so NOT true for all numbers.
For subtraction to be associative, it would have to be true that "a- (b- c)= (a- b)- c" for all numbers, a, b, c. But 3- (4- 1)= 3- 3= 0 and (3- 4)- 1= (-1)- 1= -2 shows that is NOT true for those numbers and so NOT true for all numbers.
(By the way, saying a statement is "not true for all numbers" is NOT the same as saying it is "untrue for all numbers". Do you see the difference? It is the difference between "not (true for all numbers)" and "not true (for all numbers)".
@HallsofIvy
yes I see. I am not trying to be a big ego about this. I have limited knowledge you guys are experts and know whats what. I see that there is proof by an example that commutativity and associativity are "not", "always" true for subtraction. thanks for that distinction. it has taken me some thinking to get to this. but I think I have the dragon by the tail now! it is so evil how they show this in the book! they say here is the LAW of algebra. then they break the law! I see there is a lot more to this. but they need to rewrite these books so you can actually use these laws to do algebra step by step with the laws.