Proving Field Elements Equality: a+b=a+c $\Rightarrow$ b=c

[Dansuer]

Homework Statement

Prove that if a,b,c are elements of a field.

than a+b = a+c implies b=c

Homework Equations

a + (-a) = 0

0 + a = a

The Attempt at a Solution

The solution i found is:

a+b = a+c
a+b+(-a) = a+c+(-a)
0+b = 0+c
b=c

what I'm not sure about is that since i don't know that a+b = a+c is the same as b=c, in other words, that i can subtract from both sides of an equation. can i assume that i can add to both sides of the equation?

 

[mtayab1994]

Yes because when you have something implies something else: you assume the first and you show the implication. In your case you have a+b=a+c implies b=c.

You assume a+b=a+c and you prove b=c:

You were given that 0+a=a so therefore: a+b=a+c implies a+b-(a)=c which then implies b=c

And that is correct.

 

[Dansuer]

but how do i know that a+b = a+c implies a+b+(-a) = a+c+(-a) ?

 

[mtayab1994]

but how do i know that a+b = a+c implies a+b+(-a) = a+c+(-a) ?

Well weren't you given 0+a=a?

 

[mtayab1994]

or you can use proof by contradiction: a≠b implies a+b≠a+c

 

[Dansuer]

Well weren't you given 0+a=a?

i don't see the connection

 

[mtayab1994]

i don't see the connection

If you were given 0+a=a then you have to use it in your proof .

 

[mtayab1994]

ok to make it simpler for you: 0+a=0 means that 0+a-(a)=a-(a) then you get 0=0 which is true. now: a+b=a+c

a+b-(a)=a+c-(a) (because of what you reached ^) and when you subtract a from both sides you don't change anything. This is really simple proof, nothing too complicated.

 

[mallesh.koram]

a+b+(-a-b)=0 [from a+(-a)=0]
a+c+(-a-b)=0 [from a+b=a+c]
a+c-a-b=0
c-b=0
c=b.

 

[Dansuer]

ok to make it simpler for you: 0+a=0 means that 0+a-(a)=a-(a) then you get 0=0 which is true.

you added -(a) on the left and a+(-a) on the right.

 

[mallesh.koram]

am adding in the right side only, am not gettin , where i take on left?

 

[Dansuer]

am adding in the right side only, am not gettin , where i take on left?

i was not talking to you :)

i like your proof but still

c-b = 0
c = b

is not obvious

 

[Fredrik]

but how do i know that a+b = a+c implies a+b+(-a) = a+c+(-a) ?

The answer is actually extremely trivial. The equality sign in a+b=a+c means that a+b and a+c represent exactly the same thing, i.e. the same member of the field. So what you're asking is equivalent to asking if x=x implies x+y=x+y. Yes, it's that simple.

This doesn't mean that I think it was a bad question. A lot of people don't see this right away. So you shouldn't feel bad about not seeing it. Instead you should try to remember that you would have seen it if you had only asked yourself what the notation you're looking at really means. You will encounter many problems in the future that will seem difficult at first, but will be really easy (or at least significantly easier) if you just ask yourself that same question.

By the way, the solution will be slightly simpler if you add -a from the left instead of from the right. a+b=a+c implies -a+(a+b)=-a+(a+c). Now use that addition is associative. Then use that -x+x=0 for all x. And finally, use that 0+x=x for all x. (If you start by adding -a from the right, then there are two extra steps: Use that addition is commutative and use that addition is associative one more time: (a+c)+(-a)=a+(c+(-a))=a+((-a)+c)=(a+(-a))+c=0+c=c).

Edit: Another "by the way": I think that until you have gotten really used to doing these proofs, it's a good idea to write out exactly what you're doing at each step, as I did for another problem here: (I avoided the term "field" because that guy was clearly not familiar with it).

I would take these assumptions as the starting point:

ℝ is a set.
0 and 1 are members of ℝ.
Addition and multiplication are both functions from ℝ×ℝ into ℝ.
For each x in ℝ, there's a member of ℝ denoted by -x.
For each x in ℝ except 0, there's a member of ℝ denoted by x^-1.

For all x,y,z in ℝ,

(1) (x+y)+z=x+(y+z)
(2) x+0=0+x=0
(3) x+(-x)=(-x)+x=0
(4) x+y=y+x
(5) (xy)z=x(yz)
(6) x1=1x=x
(7) xx^-1=x^-1x=1
(8) xy=yx
(9) x(y+z)=xy+xz
(10) (x+y)z=xz+yz

Then I would state and prove the theorem like this:

Theorem: For all x in ℝ, x0=0.

Proof: Let x be an arbitrary member of ℝ. Axiom (2) implies that 0=0+0.
\begin{align}
& x0=x(0+0) &&\text{Use axiom (9).}\\
& x0=x0+x0 && \text{Add $-x0$ to both sides.}\\
& x0+(-x0)=(x0+x0)+(-x0) && \text{Use axioms (3) and (1).}\\
& 0=x0+(x0+(-x0)) &&\text{Use axiom (3).}\\
& 0=x0+0 &&\text{Use axiom (2).}\\
& 0=x0
\end{align}

 

[Dansuer]

I probably explained myself not clearly, English is not my main language so please forgive me :)

x = x implies x+y = x+y that's exactly what i want to know if it's true.

i know it's true for real numbers, i learned it in elementary school. I know it's true even for complex numbers. But now I'm starting to deal with abstract objects where the operations are not the usual ones.
They asked me to prove that a+b = a+c implies b=c which in words is the old "subtract 'a' from both sides". That means that nothing can be taken for granted in those abstract stuff, not even a rule that even 12 years old children had master. Now I'm asking myself if i can "add a to both sides" namely x=x implies x+y = x+y. can it be proven? or it's an axiom? or...?

 

[Fredrik]

It follows from the meaning of the equality sign. An implication "A implies B" is false if and only if A is true and B is false. So the statement "for all x and y, x=x implies x+y=x+y" can only be false if there's a choice of x and y such that x+y≠x+y, and how can anything not be equal to itself?

 

[Dansuer]

awesome! thanks a lot!