How do you think about Subtraction?

[PhysicsHelp12]

The people on yahoo answers confused me ...and said I was doing it wrong


When you do algebra...you're supposed to think about it in terms of an additive inverse

(as multiplying by (-1) and adding it ) so you get communitivity and associativity, right?


Is that how you think of it -in terms of addition when doing algebra?

 

[gunch]

How exactly do you mean think? Do you want the definition usually given, or the intuitive concept of subtraction. Normally we take it as an axiom that every x has an additive inverse y such that x+y=0, and we then think of subtraction as the addition of the additive inverse. Or in more ordinary terms we define -x to mean the additive inverse of x, and define x-y to be equal to x+(-y). Note that you would have to prove the uniqueness of the additive inverse for subtraction to be well-defined, but that's fairly trivial.

Of course all correct definitions are equal, so you can define it however you want. If you have defined -1 (the additive inverse of the identity), and multiplication between arbitrary elements then you can define x-y to mean x+(y*-1) and it would be just as valid as the first definition I gave (in fact they can be proven equivalent for most definitions of multiplication).

The problem with you approach is that for some groups we haven't defined a multiplication operator, we have defined several multiplication operators, or it doesn't behave like a normal multiplication operator. However for fields and rings the two definitions are equivalent.

I just realized that you may be only thinking of elementary algebra. If you are only working within the real or complex numbers then the definitions are equivalent as well, so yours isn't inferior in any sense.

 

[HallsofIvy]

Both subraction and division are not very good operations- in particular they are not 'associative'. It is much better to think of them just as adding the "additive inverse" and multiplying by the "multiplicative inverse".

 

[tiny-tim]

Hi PhysicsHelp12!

Subtracting something is adding the inverse of it.

(and inverse can be defined without using multiplication, of course)

But …

The people on yahoo answers
… ...said I was doing it wrong

… how were you doing it?

 

[PhysicsHelp12]

My problem is ...I'm in university and I've been doing algebra for years now ...

so usually the right answer pops into my head without much conscious thought

but I am kind of getting worried/obsessive about 'What if I forget someday if it always

just comes to me without actually thinking much about it"

so I've been trying to think of what rules exactly I am applying --


but then there's trivial ones like associativity or commutivity that I don't think about


so Is it ok to just --Let the answer cometo you from experience --without actually

THinking about the way the rule was initially presented in its simplest form if you know its right


Eg. 1/x=x^(-1) was the basic rule


but then you apply it to a rational expression ...then technically youre using associativity ...i mean ...whats the process

ab/(cd)=ab/c*d^(-1)
...but youre using associativity arent you?
a*b/(c*d)=a*b/c*(1/d) =a*b/c*(d^(-1))=a*b/c*d^(-1)

Do you see my frustration --trying to do it completely axiomatically ..with associativity

and then it takes 10 minutes --

even though I know by experience

idk ...

Analysis and Abstract algebra have made me crazy I think...

 

[slider142]

Apply the rules to check your answer or provide a proof of your intuition to other mathematicians. Intuition and foresight is just as valuable as being able to provide a proper proof. Sometimes your intuition will be wrong, usually because you're dealing with a mathematical object you have not had past experience with, or dealt with at the same depth of analysis. Attempting to prove the statement your intuition makes will then show you your error and beef up your experience.

 

[tiny-tim]

… so I've been trying to think of what rules exactly I am applying --

but then there's trivial ones like associativity or commutivity that I don't think about
…
Eg. …
ab/(cd)=ab/c*d^(-1)
...but youre using associativity arent you?
a*b/(c*d)=a*b/c*(1/d) =a*b/c*(d^(-1))=a*b/c*d^(-1)

Do you see my frustration --trying to do it completely axiomatically ..with associativity

and then it takes 10 minutes --

It's a ball-park thing …

just remember which ball-park you're in! ​

Once you're happy that associativity and commutativity apply, then yes you can forget about them.

Don't worry! ​

 

[symbolipoint]

The people on yahoo answers confused me ...and said I was doing it wrong


When you do algebra...you're supposed to think about it in terms of an additive inverse

(as multiplying by (-1) and adding it ) so you get communitivity and associativity, right?


Is that how you think of it -in terms of addition when doing algebra?

Actually I think of subtraction both as the subtraction operation AND as the additive inverse; switching between the two ways of thinking is not difficult (is this just ME and a few other people, or is this ME and most other people?). I use the form which is most convenient at the time.

 

[PhysicsHelp12]

Ok that makes a lot of sense Thx Tiny-Tim :)