Help with proofs...

[daon]

I need help with a couple proofs. First, my teacher is saying that the following proof is accurate, and I don't see why:

For all integers m and x,
If m+x=m, x has the unique property that x=0
PF:
Let m = 0, and substitute into the original equation for m;
0+x=0. Substitute in x for [0+x] (we already proved that 0+x=x),
Now we have x=0.

My question is, how does this prove that it is true for ALL m if we only used 0? I have already done a proof, albiet more complicated, that works, but I dont see how this one works for ALL integers m.

Also, I am having trouble with the proof that for integers m and x,
If m*x=m, then x has the property x=1 for some m.

If I prove something to be true for all of a set, should I then even have to consider that it is true for some it? A lot of problems I have to do first ask to prove that somehting is true for all integers and then again for some integers. I am having a lot of trouble with proofs and would appreciate some direction as well. Are there any websites that I might find useful? I already took discrete and have done lots of proofs, but never had to use as much detail as I am required of in this class.

 

[pka]

It is next to impossible to comment on a given proof with having a complete set of axioms to work from. However, in this case you might note that the statement is better put this way: “If there is an integer x such that for every integer m, m+x=n then x has the unique zero property.” Assuming you have an axiom or a theorem which states that 0 is unique, then the proof works.
Here is why. If m+x=m is true for every m then it is true for m=0. Hence x=0.
But to be sure, we need all the axioms and theorems that precede the problem.

 

[daon]

Well, so far we have defined 0 and 1 as the only unique elements. The axioms we have defined are:

1)the commutative, associative laws for addition and multiplication as well as the distributive law
2)m+0=m for all integers m
3) 1 != 0 and m*1=m for all integers m
4) for each integer m, m+x=0 has a solution (additive inverse) which is unique, x=-m
5) for all integers m,n,p, if m!=0 and mn=mp, then n=p

So far thats all we have.We have also proved that m*1=m, and 0+m=m.

In your explanation of the previous proof, you don't have to take into account all m? I guess since you temporarily take the "If" part of the statement as true, then you can choose 0 as a substitute for m? Just a little confusing to me...

 

[ChaoticLlama]

This is the proof my Algebra presented to us this year.

Neutral Element of Vector Addition:

If \(\displaystyle \vec u \in V\)
Then \(\displaystyle \vec v + \vec u = \vec v\)

Let \(\displaystyle \vec v = \vec {AB}\)

\(\displaystyle \vec {AB} = \vec u + \vec {AB}\)

\(\displaystyle \vec {AB} = \vec {BB} + \vec {AB}\)

Therefore \(\displaystyle \vec {BB}\) is the neutral vector, where \(\displaystyle \vec {BB} = \vec 0\) such that it has magnitude zero units and undefined direction.

 

[Matt]

I need help with a couple proofs. First, my teacher is saying that the following proof is accurate, and I don't see why:

For all integers m and x,
If m+x=m, x has the unique property that x=0
PF:
Let m = 0, and substitute into the original equation for m;
0+x=0. Substitute in x for [0+x] (we already proved that 0+x=x),
Now we have x=0.

My question is, how does this prove that it is true for ALL m if we only used 0? I have already done a proof, albiet more complicated, that works, but I dont see how this one works for ALL integers m.

It doesn't. I totallly agree with you; the proof falsely assumes that m is 0 (when in fact, m can be any integer, positive, negative, or 0).

To complete the proof for arbitrary integers m:

\(\displaystyle \L\qquad\begin{eqnarray}
x&=&x+m+(-m)\\
&=&m+(-m)\\
&=&0
\end{eqnarray}\)

 

[Matt]

Assuming you have an axiom or a theorem which states that 0 is unique, then the proof works. Here is why. If m+x=m is true for every m then it is true for m=0. Hence x=0.

No, that's incorrect. The statement to prove is:

. . . "For all integers m and x, if m+x=m then x=0"

not:

. . . "If m+x=m for all integers m and x, then x=0"

 

[pka]

Matt this time you simply did not understand what I wrote!
Next when you quote me quote the entire quote!
I wrote ““If there is an integer x such that for every integer m, m+x=n then x has the unique zero property.”
Then the proof does work!
I was well aware that is not what the problem said.
The difficulty with the original statement is a matter quantification.

 

[reaper_pball69]

PKA i like the quote
There are 10 kinds of people: those who understand the binary numbers and those who don’t.

 

[reaper_pball69]

I don't really think he needs the vector def'n of this, it just makes things a little more complicated.

What calss do u need this for, Pure Calculus???

 

[daon]

Sorry, I have been busy lately. I meant to come by and thank you all for the help. pka, you were correct since we must assume that m+x=m must be true for all m, so any substitution for m would be valid.

reaper, this is just a proofs class. This class is probably the most difficult class I have ever taken.

I am taking this class simultaneously with ODE and an Algorithms class, and I am devoting 95% of all my study time to this. Maybe its just the teacher. She gives no partial credit on anything and 50% of our grade is based on these assignments. I have had to resubmit proofs 2-3 times sometimes before she deemed them correct, and it is quite fusterating to find that I got a proof wrong because I wasn't specific enough about a certain operation (although my logic has always been correct). I must admit though, that It is forcing me to learn to pay more attention to detail.

This is why I am looking for a study aid. Maybe a good website or an afordable, simplistic book. Our texbook is written by a professor here at SUNY Binghamton and I find it unacceptable for the assignments we are being given. I find I am getting better at these proofs, but sometimes I don't even know where to start.