Question

Prove, in particular, that if F is a field, and if \(\alpha, \beta ~~and~~\gamma\in F\), then the following relations hold
a) 0 + \(\alpha =\alpha\)
b)If \(\alpha +\beta =\alpha +\gamma,~~then~~\beta=\gamma\)
c)\(\alpha+(\beta-\alpha)=\beta\)
d)\(\alpha*0=0*\alpha=0\)
e)\((-1)\alpha =-\alpha\)
f)\((-\alpha)(-\beta)=\alpha\beta\)
g)if \(\alpha\beta =0, ~~then~~either~~\alpha=0~~or~~\beta=0~~(or~~both)\)
Please, help

Answer 1

\[\alpha, \beta ~~and~~\gamma\in F, \text{then the following relations hold}\]
a) \(0 + \alpha =\alpha\)

Answer 2

Lost the connection!! :(

Answer 3

Hint: start with b). Let me use my own symbols \(x,y,z\) for that manner, I find it unpleasant to ender in greek letters. So I am not very familiar with the american notation for fields, but consider \(F\) a field with its axioms and then let \(x,y,z \in F\):
\[x+y=x+z \]
use the commutative law twice, meaning that \[ x+y=x+z \implies y+x=z+x\]
Now lets use that every element in \(F\) has an additive inverse, this is an axiom from the Field definition, so lets denote \(x'\) as the additive inverse of \(x\) meaning that \(x+x'=0\) (Check the axioms)

So we obtain \[ (y+x)+x'=(z+x)+x'\] and we use the associativity (another axiom of the field) to obtain \[y+(x+x')=z+(x+x') \implies y+0=z+0 \implies y=z \ \square \]

Answer 4

c) is a piece of cake if you think about it, it only requires commutativity and associativity

Answer 5

@Spacelimbus Thanks for guiding me. I really don't know how to start. If you can, please give me some links for me to study how to approach those kinds of questions.

Answer 6

Hmm let me think, on top of my head I don't know any adequate link. I studied this topic with the book by Jänich and Fischer, both titled Linear Algebra I and II. However, when it comes to prove such statements yourself, always keep the following in mind.

A Field is something which is defined as a mathematical context. It has certain axioms that it must fulfill. The questions you have on top is basically only a way to show that with the simple axioms (Commutativity, Associativity, Distributivity and the elements of identity of addition/multiplication and of course inverse elements of addition/multiplication) you can prove all the following statements.

So whenever you want to prove a statement yourself, make sure that it's a) not already included in the axioms, it is useless to prove an axiom, so check that it enhances/advances the list of axioms/properties and b) make use of the axioms to complete your proof.

Answer 7

That is the first lecture I read on book for next semester. I was panicked with the problems. It seems obvious and that is the most difficult in proving, :)
Ok, I'll try.

Answer 8

Here's a proof for \(\large \alpha\cdot 0 = 0\). Note that \(\large \alpha\cdot 0 = \alpha\cdot (0+0)\) and thus by distributive law, we have \(\large \alpha\cdot 0 = \alpha\cdot 0 + \alpha\cdot 0\). Adding \(\large -\alpha\cdot 0\) to both sides gives us \(\large \alpha\cdot 0=0\) and this completes the proof. You pretty much have the same argument showing that \(\large 0\cdot \alpha = 0\).

Answer 9

@ChristopherToni It is d) right?
you mean I have to break it into 2 sides. Do the same process for both and at the end, both sides leads to 0 . The proof ends there, right?
The net was so bad here