[Q] Uniqueness of solution of x+a=b from field properties
Alan E. Feldman
In my Introduction to Analysis book (Rosenlicht) it states the five
field properties thusly:
[begin quote]
Property I. (Commutitivity). For every a,b in R, we have a+b=b+a
and a*b=b*a.
Property II. (Associativity). For every a,b,c in R, we have
(a+b)+c=a+(b+c) and (ab)c = a(bc).
Property III. (Distributivity): For every a,b,c in R, we have a(b+c)
= ab + ac.
Property IV. (Existence of neutral elements). There are distinct
elements 0 and 1 of R such that for all a in R we have a+0=a and
a*1=a.
Property V. (Existence of additive and multiplicative inverses). For
any a in R there is an element of R, denoted -a, such that a +
(-a) = 0, and for any nonzero a in R there is an element of R,
denoted a^(-1), such that a*a^(-1) = 1.
[OK. Then he deduces 10 conclusions from this, F1 thru F10. F1 and F2
are simply generalizations of 1 and 2 above to sums and products of
more than two terms, so I won't quote them here. OK.]
[Now, ] F3: For any a,b in R the equation x + a = b has one and
only one solution. For if x in R is such that x + a = b, then
x = x + 0
= x + (a + (-a))
= (x + a) + (-a)
= b + (-a),
so x = b + (-a) is the only possible solution; that this is indeed a
solution is immediate. [OK so far, except how do we know that -a is
unique? But that is part of my question that follows. Please continue
reading.]
One consequence is that the element 0 of Property IV is unique;
another is that for any a in R, the element -a of Property V is
unique. [end of quote]
Now hold on just a minute. Doesn't one need to assume that 0, and
-a, are unique when starting from x = x + 0 and from there to x =
b + (-a) in order to claim that x = b + (-a) is also unique? If
zero is not unique, then -a may also not be unique, and then b + (-a)
might also not be unique. No?
So if we do need to assume uniqueness of zero and -a from the start,
then we are concluding something that we had already assumed,
resulting in circular reasoning. My apologies if I am missing
something obvious, or if I have not stated my question clearly enough,
but can someone please clarify this for me? IOW, how is this not
circular reasoning?
(NOTE: I looked at the thread about a similar question from last June
(July?) but this question was not asked in that thread, and nothing in
that thread answered this question.)
Thanks.
Alan E. Feldman
Justin Davis
equal to 0. Then 0 + e = e obviously but 0 + e = 0 also, since e is another
additive identity. So e = 0... contradiction. There is only one additive
identity. It's not assumed, but follows from the definition of an additive
identity.
Let's assume you've got additive inverses b and c (not equal) of a. Then you
have that a + b + c = (a + b) + c = 0 + c = c. However, a + b + c = a + c +
b = (a + c) + b = 0 + b = b. Again, a contradiction. there is only one
additive inverse. Again, it's not assumed, but follows from the definition
of an additive inverse.
Robert Israel
Alan E. Feldman <spamsi...@yahoo.com> wrote:
>[Now, ] F3: For any a,b in R the equation x + a = b has one and
>only one solution. For if x in R is such that x + a = b, then
> x = x + 0
> = x + (a + (-a))
> = (x + a) + (-a)
> = b + (-a),
>so x = b + (-a) is the only possible solution; that this is indeed a
>solution is immediate. [OK so far, except how do we know that -a is
>unique? But that is part of my question that follows. Please continue
>reading.]
At this point, as far as you know, there could be lots of
possible choices for -a: we pick one and call it "-a".
Then the proof says x = b + (-a) [the one we just chose].
Of course the same would have worked if we had made some other
choice. So...
>One consequence is that the element 0 of Property IV is unique;
>another is that for any a in R, the element -a of Property V is
>unique. [end of quote]
Yes, namely take b=a for the first, and b = 0 for the second.
>Now hold on just a minute. Doesn't one need to assume that 0, and
>-a, are unique when starting from x = x + 0 and from there to x =
>b + (-a) in order to claim that x = b + (-a) is also unique? If
>zero is not unique, then -a may also not be unique, and then b + (-a)
> might also not be unique. No?
No. If z is any "zero", i.e. any member of R such that z + a = a,
then F3 says that z = a + (-a) (for any choice of "-a"),
and this by definition is 0, so z = 0. If q is any "-a", i.e.
any member of R such that q + a = 0, then F3 says q = 0 + (-a) = -a.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
Bill Dubuque
>
> [Now, ] F3: For any a,b in R the equation x + a = b has one and
> only one solution. For if x in R is such that x + a = b, then
>
> x = x + 0
> = x + (a + (-a))
> = (x + a) + (-a)
> = b + (-a),
>
> so x = b + (-a) is the only possible solution; that this is indeed a
> solution is immediate. [OK so far, except how do we know that -a is
> unique? But that is part of my question that follows. Please continue
> reading.]
>
> One consequence is that the element 0 of Property IV is unique;
> another is that for any a in R, the element -a of Property V is
> unique. [end of quote]
>
> Now hold on just a minute. Doesn't one need to assume that 0, and
> -a, are unique when starting from x = x + 0 and from there to x =
> b + (-a) in order to claim that x = b + (-a) is also unique? If
> zero is not unique, then -a may also not be unique, and then b + (-a)
The proof shows that given any two roots r, r' of x + a = b
then r = b + -a, r' = b + -a, so r = r'. It matters not
what possibly non-unique value of -a or 0 the proof employs.
Any doubts dissipate if you contrast the above uniqueness proof
with the equivalent proof that r + a = r' + a => r = r'
Here too one cancels a using possibly non-unique -a and 0
but neither of them linger in the result, avoiding any confusion.
-Bill Dubuque
Leonard Blackburn
You've already gotten some useful replies. However, I am wondering why
nobody remarked that you are correct and that the author has made an error
(in my opinion it is very important to explicitly acknowledge this).
Above, you state F3 as a theorem and give the author's proof. The author's
proof is wrong since he did not first prove the uniqueness of -a (the
uniqueness of 0 doesn't play a role in his proof). You are to be commended
for paying close attention to your book.
I would have done things in this order:
1. state the five field properties.
2. show that 0 is the unique element with the property stated
(if 0' is another element with that property, then 0 = 0 + 0'
(by property IV for 0') = 0' + 0 (by property I) = 0' (by
property IV for 0).
3. show that -a is unique: suppose a + x = 0 and a + y = 0.
then x = x + 0 = x + (a + y) = (x + a) + y = (a + x) + y
= 0 + y = y + 0 = y. So, there is a unique x such that
a + x = 0, and we call it -a.
4. now prove that for any a and b the equation x + a = b has
exactly one solution. the proof that the author gave is now
valid.
Good luck,
Leonard
Exercise: show that 1 is unique. show that a^(-1) is unique.
show that the equation x*a = b has exactly one solution whenever
a is not 0.
Bill Dubuque
>> only one solution. For if x in R is such that x + a = b, then
>>
>> x = x + 0
>> = x + (a + (-a))
>> = (x + a) + (-a)
>> = b + (-a),
>>
>> so x = b + (-a) is the only possible solution; that this is indeed a
>> solution is immediate. [OK so far, except how do we know that -a is
>> unique? But that is part of my question that follows. Please continue
>> reading.]
>
> You've already gotten some useful replies. However, I am wondering why
> nobody remarked that you are correct and that the author has made an error
> (in my opinion it is very important to explicitly acknowledge this).
> Above, you state F3 as a theorem and give the author's proof. The author's
> proof is wrong since he did not first prove the uniqueness of -a (the
> uniqueness of 0 doesn't play a role in his proof).
The author has not made an error -- the proof is correct as it is.
The proof employs only the existence (not the uniqueness) of both
an additive inverse of a (denoted -a) and neutral elt (denoted 0).
Uniqueness is a corollary, the special-cases b = 0; b = a resp.
The proof is a little slick -- perhaps too much so for neophytes.
-Bill Dubuque
Alan E. Feldman
All right, let's not get into insults. I'm interested in answers. OK?
This gets to the crux of the problem. How can x = b + (-a) be known
to be unique when -a is not known to be unique? Sure, you can say
the solution has to be x = b + (-a), but if -a is not unique, then
how can x be? Suppose -a could be 3 or 5. Then x could be b
+ 3 or b + 5 and would then not be unique. So please explain how a
not-yet-known-or-shown-to-be-unique, i.e., a possibly multi-valued,
-a, could be added to b and produce a unique value for x.
Thanks
Alan E. Feldman
Alan E. Feldman
That's all fine. I don't doubt that they are unique. But the author
claims that the uniqueness of zero and the uniqueness of additive
inverses follow from the uniqueness of the solution to x = b + (-a).
Now you're saying the uniqueness follows from the properties without
the x = b + (-a) argument. My question is: How does the author
determine that x = b + (-a) is unique without assuming that zero
and/or the additive inverse of a are/is unique?
Alan E. Feldman
Alan E. Feldman
> In article <b096a4ee.03051...@posting.google.com>,
> Alan E. Feldman <spamsi...@yahoo.com> wrote:
>
> >[Now, ] F3: For any a,b in R the equation x + a = b has one and
> >only one solution. For if x in R is such that x + a = b, then
>
> > x = x + 0
> > = x + (a + (-a))
> > = (x + a) + (-a)
> > = b + (-a),
>
> >so x = b + (-a) is the only possible solution; that this is indeed a
> >solution is immediate. [OK so far, except how do we know that -a is
> >unique? But that is part of my question that follows. Please continue
> >reading.]
>
> At this point, as far as you know, there could be lots of
> possible choices for -a: we pick one and call it "-a".
> Then the proof says x = b + (-a) [the one we just chose].
> Of course the same would have worked if we had made some other
> choice. So...
But wait. If there are other choices for -a, then wouldn't they,
when added to b, produce different values for x, thereby making x
*not* unique?
> >One consequence is that the element 0 of Property IV is unique;
> >another is that for any a in R, the element -a of Property V is
> >unique. [end of quote]
>
> Yes, namely take b=a for the first, and b = 0 for the second.
>
> >Now hold on just a minute. Doesn't one need to assume that 0, and
> >-a, are unique when starting from x = x + 0 and from there to x =
> >b + (-a) in order to claim that x = b + (-a) is also unique? If
> >zero is not unique, then -a may also not be unique, and then b + (-a)
> > might also not be unique. No?
>
> No. If z is any "zero", i.e. any member of R such that z + a = a,
> then F3 says that z = a + (-a) (for any choice of "-a"),
No, I'm asking about proving F3, not deducing uniqueness of zero and
-a from it. I am asking how you can claim that x = b + (-a) is
unique when you haven't shown that zero and -a are each unique.
> and this by definition is 0, so z = 0. If q is any "-a", i.e.
> any member of R such that q + a = 0, then F3 says q = 0 + (-a) = -a.
No, no, no. I was saying that doesn't 0 and -a already have to be
known or shown to be unique *before* you can say that x = b + (-a)
is unique?
I apologize if my original post was unclear. Thanks for your response.
> Robert Israel isr...@math.ubc.ca
> Department of Mathematics http://www.math.ubc.ca/~israel
> University of British Columbia
> Vancouver, BC, Canada V6T 1Z2
Alan E. Feldman
Alan E. Feldman
OK. Let -a = 3, and let (-a)' = 5, just to be concrete about it.
You said that
r + a = r' + a . OK.
The solution is claimed to be b + (-a). So then we get
r = b + (-a) = b + 3
r' = b + (-a)' = b + 5
Then,
(b + 3) + a = (b + 5) + a' ! by subst.
b + (3 + a) = b + (5 + a') ! by assoc. law
b + (-a + a) = b + [(-a)' + a'] ! by subst.
b + 0 = b + 0 ! by additive inverses add to zero
b = b ! by zero being additive identity
yet r does not equal r'.
The doubt remains. I'm not saying that the author's proof of F3 is
wrong. I'm saying that I don't quite see it and would appreciate it if
someone would explain better how it's right or prove it wrong.
Alan E. Feldman
Alan E. Feldman
Thank you!
>
> I would have done things in this order:
>
> 1. state the five field properties.
>
> 2. show that 0 is the unique element with the property stated
> (if 0' is another element with that property, then 0 = 0 + 0'
> (by property IV for 0') = 0' + 0 (by property I) = 0' (by
> property IV for 0).
>
> 3. show that -a is unique: suppose a + x = 0 and a + y = 0.
> then x = x + 0 = x + (a + y) = (x + a) + y = (a + x) + y
> = 0 + y = y + 0 = y. So, there is a unique x such that
> a + x = 0, and we call it -a.
>
> 4. now prove that for any a and b the equation x + a = b has
> exactly one solution. the proof that the author gave is now
> valid.
>
> Good luck,
> Leonard
Looks good to me.
[...]
Alan E. Feldman
Paul Sperry
Feldman <spamsi...@yahoo.com> wrote:
Let's try it this way. Pick your favorite inverse of a - call it W. We
know a + W = I where I is an identity. Then if x + a = b,
x + a + W = b + W; x + I = b + W; x = b + W.
Now, your problem as I see it is what if I pick _my_ favorite inverse
of a, say V. Then x = b + V and, unless we know W = V we don't know x
is unique. A point well taken IMO.
--
Paul Sperry
Columbia, SC (USA)
RL
The argument you have just given says
"if x is *any*
solution to `x+a=b', then
x=b+W (*)"
so if y is *any* element with y+a=b, then
y = b+W
= x
Hence we have a unique solution. As Bill Dubuque says above, all we
needed was *some* W with W+a=0, whether this W is unique does not
matter for this proof. (what happened if we picked a different choice
for the additive inverse? well then your argument with W replaced by V
tells us that
z = b+V is *a* solution, and hence by (*), z=x.
Bill Dubuque
>>
>> Any doubts dissipate if you contrast the above uniqueness proof
>>
>> with the equivalent proof that r + a = r' + a => r = r'
>>
>> Here too one cancels a using possibly non-unique -a and 0
>>
>> but neither of them linger in the result, avoiding any confusion.
>
>
> You said that
>
> r + a = r' + a . OK.
What I mean is this: if r, r' are two solutions of x + a = b
then r + a = r' + a, since both are equal to b. Now one may
deduce r = r' by adding -a to both sides, where -a is
*any* additive inverse of a, just as in Rosenlicht's proof.
So any two solutions r, r' are equal, i.e. solutions are unique.
This approach is less subtle and less likely to cause confusion.
Indeed, the proof shows: a has an inverse => a is cancellable,
a property familiar from all the fundamental "number" systems.
Bill Dubuque
>>Leonard Blackburn <blac...@math.umn.edu> wrote:
>>>>Alan E. Feldman <spamsi...@yahoo.com> wrote:
>>>>
>>>> [Now, ] F3: For any a,b in R the equation x + a = b has one and
>>>> only one solution. For if x in R is such that x + a = b, then
>>>>
>>>> x = x + 0
>>>> = x + (a + (-a))
>>>> = (x + a) + (-a)
>>>> = b + (-a),
>>>>
>>>> so x = b + (-a) is the only possible solution; that this is indeed a
>>>> solution is immediate. [OK so far, except how do we know that -a is
>>>> unique? But that is part of my question that follows. Please continue
>>>> reading.]
>>>
>>> You've already gotten some useful replies. However, I am wondering why
>>> nobody remarked that you are correct and that the author has made an
>>> error (in my opinion it is very important to explicitly acknowledge this).
>>> Above, you state F3 as a theorem and give the author's proof. The
>>> author's proof is wrong since he did not first prove the uniqueness of
>>> -a (the uniqueness of 0 doesn't play a role in his proof).
>>
>> The author has not made an error -- the proof is correct as it is.
>> The proof employs only the existence (not the uniqueness) of both
>> an additive inverse of a (denoted -a) and neutral elt (denoted 0).
>> Uniqueness is a corollary, the special-cases b = 0; b = a resp.
>
> to be unique when -a is not known to be unique? Sure, you can say
> the solution has to be x = b + (-a), but if -a is not unique, then
> how can x be? Suppose -a could be 3 or 5. Then x could be b
> + 3 or b + 5 and would then not be unique. So please explain how a
> not-yet-known-or-shown-to-be-unique, i.e., a possibly multi-valued,
> -a, could be added to b and produce a unique value for x.
I explained this in a prior post, but perhaps I was too terse, so I
elaborate: To say that the solutions x of x + a = b are unique
means precisely: if r, r' are two solutions for x, then r = r'.
Rosenlicht's proof shows: if addition is associative with neutral 0
and a has some additive inverse -a, then *any* solution x must
satisfy x = b + -a. So if r, r' are two solutions, they must both
be equal to b + -a, hence r = r', i.e. any solutions are unique.
The structure of the proof is clarified if we abstract a little.
Above we have a set S (of solutions) which we wish to prove has
no more more than one element. To prove this it suffices to prove
there *exists* an "equalizer" e for S, i.e. x in S => x = e
Rosenlicht's proof shows that e = b + -a is an equalizer for the
set S = {x : x + a = b}. It doesn't matter that the proof made
some arbitrary choice while constructing e, since to equalize
the set S all we require is the *existence* of one such e.
>> The proof is a little slick -- perhaps too much so for neophytes.
>
> All right, let's not get into insults. I'm interested in answers. OK?
No offense was intended. I merely meant to express the opinion that,
in a textbook [*Introduction* to Analysis] intended for beginners,
it is perhaps not the best choice to present proofs involving
subtleties that lead down tangents, when such subtleties can easily
be eliminated by restructuring the proof - without increasing the
overall complexity of the proof. Of course it is a very difficult
task for the author to recognize these subtle points since his
level of knowledge is much deeper by now, and it is difficult to
"regress" and try to recall how the mathematical world looked from
a more naive standpoint. To do such requires conscious effort to
forget knowledge that by now has become subconscious and intuitive.
Such deeper knowledge may actually force a particular viewpoint
in the authors mind that is much beyond the level of the targeted
student. A good analogy here are the well known optical illusions
where an image can be interpreted in more than one way but, because
of context, one of the interpretations is dominant, and it is
sometimes very difficult to shift to the other viewpoint. The same
sort of ambiguity occurs in mathematics, only it can be much more
extreme since there are often many different views of objects
employing a wealth of abstractions far beyond any other science.
For example, suppose an author is an expert in algebraic geometry
and he's writing an introductory textbook in algebra (excluding
any geometry). He might mistakenly omit some details that are
non-trivial from the algebraic standpoint but which are trivial
when viewed geometrically. However, to recognize this the author
would need to suppress his by now innate geometric intuition
and force himself to think in the more rarefied algebraic world.
Just as with said optical illusions, this may require extreme
mental effort to continually shift from the dominant viewpoint.
-Bill Dubuque
Leonard Blackburn
You're correct and I spoke without enough thinking. Sorry.
Pedagogically speaking, I think the approach I outlined would be better,
though. More students would understand.
A further attempt to explain to the original poster why the author's
proof is correct:
While the uniqueness of (-a) has not been established before the proof
(only after, as a corollary), it isn't needed. Here's why:
1. Choose some additive identity and call it 0.
2. Given an element a, choose some element y such that a + y = 0.
Let -a denote y.
3. Suppose x satisfies x + a = b. Then, the steps the author gave
show, without a doubt, that x must be -a + b. So EVERY solution
of this equation must be -a + b, which is a fixed element since
we have fixed -a. Thus, the solution (which we can check) is indeed
unique. We didn't need the uniqueness of -a. If it were possible
to choose a different element, say a', such that a + a' = 0, then
yes, x = a' + b. But x would ALSO equal -a + b. Getting x = -a + b
and x = a' + b doesn't show that x may not be unique, it only shows
that -a + b = a' + b!
Actually, now that I think about it, the author's method may be pedagogically
better than the one I outlined earlier, because once a student forces
himself to understand it, he may be a little bit smarter.
I apologize to the author in question and to anyone else in this thread
whom I mistakenly said was wrong.
-Leonard
Leonard Blackburn
Your problem probably stems from your use of the word "or" in the preceding
sentence. Change it to "and." In fact, you should change that last
sentence to "Then x MUST be b + 3 AND b + 5." Your argument does not
show that x can possibly have different values, b + 3 or b + 5, it only
shows that b + 3 = b + 5. Let me reiterate: one can show that x = b + 3
(in your example), and then turn around and show that x = b + 5, too.
But, if you choose the inverse "5" this does not mean that x no longer
has to be b + 3. It's still b + 3 too. any solution x must still be
b + 3.
I know it's confusing. I was confused by it at first, too, even though
I have completed undergraduate and graduate courses in abstract algebra
(with grades of "A") several years ago. But let me reassure you that the
original proof you posted is correct. The uniqueness of x has been established
before knowing about the uniqueness of -a.
Other posters are throwing in one more step to the proof for you to make it
clear: if x_1 and x_2 are both solutions of x + a = b, and if we have chosen
a specific 0 and a specific -a (even if there are lots to choose from), then
x_1 = -a + b and x_2 = -a + b. So, x_1 = x_2. Thus all solutions are equal
to each other. Thus there is only one solution. This makes it clear that
the uniqueness of -a is not needed. However, even without making this
last step explicit, the author's original proof is correct AND complete.
One last attempt to convince you:
Your objection:
1. Pick an additive identity and call it 0.
2. Now suppose a + a_1 = 0, a + a_2 = 0, and a + a_3 = 0 (you are correct to
not assume that there is only one additive inverse at this stage, so for
the sake of argument, let's suppose there are possibly 3 of them).
3. Now, taking the same steps in solving x + a = b the author did, but
with using each of a_1, a_2, a_3 as possible additive inverses of a, we
would get
x = a_1 + b,
x = a_2 + b,
x = a_3 + b.
This seems to suggest to you, that there might be three solutions of the
equation instead of just one, as promised. But all this really should
suggest to you is that
x = a_1 + b = a_2 + b = a_3 + b,
not contradicting uniqueness at all. Remember that before solving the
equation, you are "fixing" x. You are saying, suppose x is an element of the
field such that x + a = b. You are not leaving x variable and amorphous.
Thus, for your supposed x, x must be each of a_1 + b, a_2 + b, a_3 + b.
It's not just one of those elements, it's ALL of them. In particular, every
x such that x + a = b must be a_1 + b (I could have said a_2 + b as well,
or a_3 + b). EVERY solution must be a_1 + b. Therefore, every solution
must be equal to every other solution. Therefore, the solution is
unique (assuming it exists, but we really don't have a problem with the
existence part).
Still, you are to be commended for questioning and looking at things closely.
I sincerely hope this post helped.
Good luck,
Leonard
Robert B. Israel
> > In article <b096a4ee.03051...@posting.google.com>,
> > Alan E. Feldman <spamsi...@yahoo.com> wrote:
> >
> > >[Now, ] F3: For any a,b in R the equation x + a = b has one and
> > >only one solution. For if x in R is such that x + a = b, then
>
> > > x = x + 0
> > > = x + (a + (-a))
> > > = (x + a) + (-a)
> > > = b + (-a),
>
> > >so x = b + (-a) is the only possible solution; that this is indeed a
> > >solution is immediate. [OK so far, except how do we know that -a is
> > >unique? But that is part of my question that follows. Please continue
> > >reading.]
> >
> > At this point, as far as you know, there could be lots of
> > possible choices for -a: we pick one and call it "-a".
> > Then the proof says x = b + (-a) [the one we just chose].
> > Of course the same would have worked if we had made some other
> > choice. So...
>
> But wait. If there are other choices for -a, then wouldn't they,
> when added to b, produce different values for x, thereby making x
> *not* unique?
The point is that we have a proof of x = b + (-a). If (-a)' is
another choice
for -a, then also x = b + (-a)'. And then b + (-a) = b + (-a)':
things equal
to the same thing are equal to each other.
> > >One consequence is that the element 0 of Property IV is unique;
> > >another is that for any a in R, the element -a of Property V is
> > >unique. [end of quote]
> >
> > Yes, namely take b=a for the first, and b = 0 for the second.
> > >Now hold on just a minute. Doesn't one need to assume that 0, and
> > >-a, are unique when starting from x = x + 0 and from there to x =
> > >b + (-a) in order to claim that x = b + (-a) is also unique? If
> > >zero is not unique, then -a may also not be unique, and then b + (-a)
> > > might also not be unique. No?
> >
> > No. If z is any "zero", i.e. any member of R such that z + a = a,
> > then F3 says that z = a + (-a) (for any choice of "-a"),
>
> No, I'm asking about proving F3, not deducing uniqueness of zero and
> -a from it. I am asking how you can claim that x = b + (-a) is
> unique when you haven't shown that zero and -a are each unique.
Let's give them different names. Suppose Z is any "zero", so in
particular
x + Z = x. Suppose A is any "-a" for this "zero", so a + A = Z. Then
the
proof says if x + a = b then x = x + Z = x + a + A = b + A. _Every_ x
that satisfies x + a = b must be equal to b + A. All those x's are
equal to the
same thing, so there's only one of them.
Alan E. Feldman
Why "AND"? If we have the equation (x-a)(x-b)=0, does x = a and x = b?
No. There are two distinct roots: a and b. I don't see why it's
different for x=b+(-a)_1, x=b+(-a)_2, where (-a)_1 and (-a)_2 are two
possible inverses of a. Actually, I believe I see the reason for AND
below; please continue reading.
> show that x can possibly have different values, b + 3 or b + 5, it only
> shows that b + 3 = b + 5. Let me reiterate: one can show that x = b + 3
> (in your example), and then turn around and show that x = b + 5, too.
> But, if you choose the inverse "5" this does not mean that x no longer
> has to be b + 3. It's still b + 3 too. any solution x must still be
> b + 3.
>
> I know it's confusing. I was confused by it at first, too, even though
> I have completed undergraduate and graduate courses in abstract algebra
> (with grades of "A") several years ago. But let me reassure you that the
> original proof you posted is correct. The uniqueness of x has been established
> before knowing about the uniqueness of -a.
>
> Other posters are throwing in one more step to the proof for you to make it
> clear: if x_1 and x_2 are both solutions of x + a = b, and if we have chosen
> a specific 0 and a specific -a (even if there are lots to choose from), then
> x_1 = -a + b and x_2 = -a + b. So, x_1 = x_2. Thus all solutions are equal
> to each other. Thus there is only one solution. This makes it clear that
> the uniqueness of -a is not needed. However, even without making this
> last step explicit, the author's original proof is correct AND complete.
But one can show that x_3 = (-a)_3 + b. Why does that have to be the
same value of x? Actually, ..., wait a minute, ... ..., I think I
see it now. This now makes sense to me. Any solutions x_n can be used
with the "original" zero and "original" inverse and thus have to equal
b+(-a). Yes, that makes sense to me now. Thank you. But there is still
more below.
> One last attempt to convince you:
> Your objection:
> 1. Pick an additive identity and call it 0.
>
> 2. Now suppose a + a_1 = 0, a + a_2 = 0, and a + a_3 = 0 (you are correct to
> not assume that there is only one additive inverse at this stage, so for
> the sake of argument, let's suppose there are possibly 3 of them).
>
> 3. Now, taking the same steps in solving x + a = b the author did, but
> with using each of a_1, a_2, a_3 as possible additive inverses of a, we
> would get
> x = a_1 + b,
> x = a_2 + b,
> x = a_3 + b.
> This seems to suggest to you, that there might be three solutions of the
> equation instead of just one, as promised. But all this really should
> suggest to you is that
>
> x = a_1 + b = a_2 + b = a_3 + b,
>
> not contradicting uniqueness at all. Remember that before solving the
> equation, you are "fixing" x. You are saying, suppose x is an element of the
> field such that x + a = b. You are not leaving x variable and amorphous.
Here is the crux. "... you are 'fixing' x". Doesn't that mean you are
assuming it is unique already at this stage?
Actually, I can see it this way:
x = x + a + (-a)
At this point, we can use what could possibly be different inverses of
a and doing so leaves the value of x unchanged. Then, by replacing
x + a with b,
x = b + (-a)
and since this is the same x that is unchanged by possibly different
values of (-a), this is "the only possible solution". And the rest
I've understood from the outset.
I'm not 100% certain, but I think this way of looking at it does not
assume that (-a) is unique (even though one can very quickly conclude
that it is). If so, then I now understand the proof of F3, at least in
this way. Hurray! (Actually, also in the way where I paused up above.)
I guess that even if you still assume different possible values for
(-a), then
b + any of those values
would still equal the same value of x. Then you have proved F3 and
can quickly set b equal to 0 to show that (-a) is indeed unique.
And, of course, you can set b equal to a to show that 0 is
unique.
OK. I think I've got it now. Yes, I believe that's what all of the
responders have been telling me, but I couldn't get it until I saw
that different inverses do not change the value of x in x=x=a+(-a)
and hence also in x=b+(-a) as I described above (and later also the
paragraph with the pause in it). My apologies for being slow about
this, but F3 is a very important conclusion based on the field
properties (because several other conclusions are based on F3) and I
really wanted to be sure I understood the original author's proof
properly.
So in summary, any x that solves x + a = b that is dervied from
any possible value of zero and any possible value of the inverse of a
can simply be used in the original derviation using the original zero
and the original additive inverse of a, and so must be equal to that
original b + (-a) and hence be unique. OK!
Thank you for your help and patience.
And thanks to all the other posters in this thread for their help.
> Thus, for your supposed x, x must be each of a_1 + b, a_2 + b, a_3 + b.
> It's not just one of those elements, it's ALL of them. In particular, every
> x such that x + a = b must be a_1 + b (I could have said a_2 + b as well,
> or a_3 + b). EVERY solution must be a_1 + b. Therefore, every solution
> must be equal to every other solution. Therefore, the solution is
> unique (assuming it exists, but we really don't have a problem with the
> existence part).
>
> Still, you are to be commended for questioning and looking at things closely.
Thank you.
> I sincerely hope this post helped.
Yes, it did.
> Good luck,
> Leonard
[...]
Alan E. Feldman
Leonard Blackburn
The situation is not analogous.
Here's a proof of why AND?
Suppose a + a_1 = 0 and a + a_2 = 0.
Suppose x is an element of the field such that x + a = b.
Then, following the steps the author gave you with a_1
replacing -a we get x = a_1 + b. Also, following the steps
the author gave you with a_2 replacing -a we get x = a_2 + b.
So, this shows that x = a_1 + b AND x = a_2 + b.
> No. There are two distinct roots: a and b. I don't see why it's
> different for x=b+(-a)_1, x=b+(-a)_2, where (-a)_1 and (-a)_2 are two
> possible inverses of a. Actually, I believe I see the reason for AND
> below; please continue reading.
Ok. But the situation IS different.
>
>
> > show that x can possibly have different values, b + 3 or b + 5, it only
> > shows that b + 3 = b + 5. Let me reiterate: one can show that x = b + 3
> > (in your example), and then turn around and show that x = b + 5, too.
> > But, if you choose the inverse "5" this does not mean that x no longer
> > has to be b + 3. It's still b + 3 too. any solution x must still be
> > b + 3.
> >
> > I know it's confusing. I was confused by it at first, too, even though
> > I have completed undergraduate and graduate courses in abstract algebra
> > (with grades of "A") several years ago. But let me reassure you that the
> > original proof you posted is correct. The uniqueness of x has been established
> > before knowing about the uniqueness of -a.
> >
> > Other posters are throwing in one more step to the proof for you to make it
> > clear: if x_1 and x_2 are both solutions of x + a = b, and if we have chosen
> > a specific 0 and a specific -a (even if there are lots to choose from), then
> > x_1 = -a + b and x_2 = -a + b. So, x_1 = x_2. Thus all solutions are equal
> > to each other. Thus there is only one solution. This makes it clear that
> > the uniqueness of -a is not needed. However, even without making this
> > last step explicit, the author's original proof is correct AND complete.
>
>
> But one can show that x_3 = (-a)_3 + b. Why does that have to be the
> same value of x? Actually, ..., wait a minute, ... ..., I think I
> see it now. This now makes sense to me. Any solutions x_n can be used
> with the "original" zero and "original" inverse and thus have to equal
> b+(-a). Yes, that makes sense to me now. Thank you. But there is still
> more below.
Glad you see it.
>
>
> > One last attempt to convince you:
> > Your objection:
> > 1. Pick an additive identity and call it 0.
> >
> > 2. Now suppose a + a_1 = 0, a + a_2 = 0, and a + a_3 = 0 (you are correct to
> > not assume that there is only one additive inverse at this stage, so for
> > the sake of argument, let's suppose there are possibly 3 of them).
> >
> > 3. Now, taking the same steps in solving x + a = b the author did, but
> > with using each of a_1, a_2, a_3 as possible additive inverses of a, we
> > would get
> > x = a_1 + b,
> > x = a_2 + b,
> > x = a_3 + b.
> > This seems to suggest to you, that there might be three solutions of the
> > equation instead of just one, as promised. But all this really should
> > suggest to you is that
> >
> > x = a_1 + b = a_2 + b = a_3 + b,
> >
> > not contradicting uniqueness at all. Remember that before solving the
> > equation, you are "fixing" x. You are saying, suppose x is an element of the
> > field such that x + a = b. You are not leaving x variable and amorphous.
>
>
> Here is the crux. "... you are 'fixing' x". Doesn't that mean you are
> assuming it is unique already at this stage?
No, it doesn't mean that. The original problem was to prove that there
is exactly one x such that x + a = b. To do this, we suppose that x is
an arbitrary element of the field such that x + a = b. We don't know
what it is yet, but we have fixed it. This is a VERY common practice in
mathematics. We want to know something about every possible x that
satisfies x + a = b. So, we choose an arbitrary x that satisfies
x + a = b, and we find out certain properties about it. In this case
we find out that x = -a + b. So, no matter which x we may have "fixed"
it must equal -a + b.
This is how a mathematician normally proves a statement beginning with
"for all x ...." He chooses, arbitrarily, an x and begins to prove
what he needs for that x. Then since it was an arbitrary choice, he
has proven what he needs for ALL x.
>
> Actually, I can see it this way:
>
> x = x + a + (-a)
>
> At this point, we can use what could possibly be different inverses of
> a and doing so leaves the value of x unchanged.
Yes.
> Then, by replacing
> x + a with b,
>
> x = b + (-a)
>
> and since this is the same x that is unchanged by possibly different
> values of (-a), this is "the only possible solution". And the rest
> I've understood from the outset.
Good.
>
> I'm not 100% certain, but I think this way of looking at it does not
> assume that (-a) is unique (even though one can very quickly conclude
> that it is). If so, then I now understand the proof of F3, at least in
> this way. Hurray! (Actually, also in the way where I paused up above.)
Good.
>
> I guess that even if you still assume different possible values for
> (-a), then
>
> b + any of those values
>
> would still equal the same value of x. Then you have proved F3 and
> can quickly set b equal to 0 to show that (-a) is indeed unique.
> And, of course, you can set b equal to a to show that 0 is
> unique.
Good.
>
> OK. I think I've got it now. Yes, I believe that's what all of the
> responders have been telling me, but I couldn't get it until I saw
> that different inverses do not change the value of x in x=x=a+(-a)
> and hence also in x=b+(-a) as I described above (and later also the
> paragraph with the pause in it). My apologies for being slow about
> this, but F3 is a very important conclusion based on the field
> properties (because several other conclusions are based on F3) and I
> really wanted to be sure I understood the original author's proof
> properly.
No need for apologies. Just glad you got it.
Just keep reading with skepticism and attention to detail. If you
keep on with math, in a couple of years you'll look back at this
problem and you won't see any of the confusion you were having before.
Good luck,
Leonard
Bill Dubuque
>
> Actually, I can see it this way:
>
> x = x + a + (-a)
>
> At this point, we can use what could possibly be different inverses of
> a and doing so leaves the value of x unchanged. Then, by replacing
> x + a with b,
>
> x = b + (-a)
>
> and since this is the same x that is unchanged by possibly different
> values of (-a), this is "the only possible solution". And the rest
> I've understood from the outset.
This is equivalent to what I've been saying all along. Namely, suppose
that x, X are both solutions of x + a = b. One immediately deduces
that x = X by canceling a from x + a = X + a, where canceling may
be achieved by addition of any inverse -a. The equivalence is as follows
x = x + a + -a = b + -a = X + a + -a = X
e.g. -a = -a + a + -a = 0 + -a = -A + a + -a = -A
A slick way to represent such derivations is via two-way evaluations:
Inverses Unique: ------
-a + a + -A since over/underlined = 0
------
=> -a = -A
Law of Signs: -----------
a b + a(-b) + (-a)(-b) since over/underlined = 0
---------------- by the distributive law
=> a b equals (-a)(-b)
-Bill Dubuque
Alan E. Feldman
> Alan E. Feldman <spamsi...@yahoo.com> wrote:
> >
> > Actually, I can see it this way:
> >
> > x = x + a + (-a)
> >
> > At this point, we can use what could possibly be different inverses of
> > a and doing so leaves the value of x unchanged. Then, by replacing
> > x + a with b,
> >
> > x = b + (-a)
> >
> > and since this is the same x that is unchanged by possibly different
> > values of (-a), this is "the only possible solution". And the rest
> > I've understood from the outset.
>
> This is equivalent to what I've been saying all along. Namely, suppose
>
> that x, X are both solutions of x + a = b. One immediately deduces
>
> that x = X by canceling a from x + a = X + a, where canceling may
>
> be achieved by addition of any inverse -a. The equivalence is as follows
>
> x = x + a + -a = b + -a = X + a + -a = X
>
> e.g. -a = -a + a + -a = 0 + -a = -A + a + -a = -A
Yes, but for some reason I didn't see it at that point. I kept
focusing on, "But if (-a) could be different, how could that not
change x?" Part of the problem was that often I would say to myself
"Yes, I think I see it ..." and then somehow it would slip away.
When I realized the fact that "other possible (-a)'s" wouldn't change
x in x=x+a+(-a) it was a breakthrough. And then I saw that any x could
still be run through the original argument to equal b+(-a) and this
could only equal that. And it all happens so fast. x=b+(-a),x
unique,(-a) unique that combined with lack of sleep I had trouble
sorting it all out. Also, I had trouble simply trying to forget that
subtraction of real numbers of course yields unique answers, and then
trying too hard to pretend that I only knew the field properties and
F1 and F2. In fact, it seems so obvious that subtraction of real
numbers gives unique answers that it is difficult to think that we
even have to prove this! Of course the point is to deduce it from
field properties, which I like seeing.
I see now that it is only the existence of the additive inverse that
is necessary for the proof.
The rest of the chapter all made sense to me, but for some reason, I
just got stuck on this one point. Hey, this book is from a math course
I took in college 20 years ago and I haven't touched it since until
now. So I can say I'm out of practice! (I majored in physics, not
math, but I like both.)
Anyway, I'm glad it's clear to me now, and I have you and the other
responders to thank for it.
> A slick way to represent such derivations is via two-way evaluations:
>
> Inverses Unique: ------
> -a + a + -A since over/underlined = 0
> ------
>
> => -a = -A
>
> Law of Signs: -----------
> a b + a(-b) + (-a)(-b) since over/underlined = 0
> ---------------- by the distributive law
>
> => a b equals (-a)(-b)
>
> -Bill Dubuque
Yes, that's neat.
Alan E. Feldman
Alan E. Feldman
> Bill Dubuque <w...@nestle.ai.mit.edu> wrote in message news:<y8ziss5...@nestle.ai.mit.edu>...
> > Alan E. Feldman <spamsi...@yahoo.com> wrote:
> I see now that it is only the existence of the additive inverse that
> is necessary for the proof.
>
> The rest of the chapter all made sense to me, but for some reason, I
> just got stuck on this one point. Hey, this book is from a math course
> I took in college 20 years ago and I haven't touched it since until
Uh, make that 23 years ago, give or take a year.
> now. So I can say I'm out of practice! (I majored in physics, not
> math, but I like both.)
>
[...]
Alan E. Feldman