Lack of understanding in argument about non-zero, non-parallel vectors
Albert
I'm trying to teach and introduce myself to vectors. I'm reading the
following:
_Linear combinations of non-parallel vectors_
If a and b are non-zero, non-parallel vectors, then
ma + nb = pa + qb
implies m = p and n = q
An argument is as follows:
ma - pa = qb - nb
(therefore) (m - p)a = (q - n)b
(therefore) a = ((q - n) / (m - p)b or b = ((m - p) / (q - n))a
but a and b are not parallel and not zero
(therefore) q = n and m = p
It's the last two lines that I don't follow (but a and b ... m = p).
What is going on here?
TIA,
Albert
Virgil
Albert <albert.xt...@gmail.com> wrote:
> Hello,
>
> I'm trying to teach and introduce myself to vectors. I'm reading the
> following:
>
> _Linear combinations of non-parallel vectors_
> If a and b are non-zero, non-parallel vectors, then
>
> ma + nb = pa + qb
>
> implies m = p and n = q
>
> An argument is as follows:
> ma - pa = qb - nb
> (therefore) (m - p)a = (q - n)b
> (therefore) a = ((q - n) / (m - p)b or b = ((m - p) / (q - n))a
Note that a = ((q - n) / (m - p) b makes no sense when m = p and
similarly b = ((m - p) / (q - n))a makes no sense when q = n, since
division by zero is not defined or allowed.
But if either equation does make sense (by not dividing by zero) then
you have two non-parallel vectors which must be parallel.
Thus each of those necessarily senseless equations must have been
constructed erroneously by dividing by zero.
Jon Slaughter
if you assume q-n != 0 and m-p != 0 then you get a = s*b which implies a is
parallel to b but we know they aren't by hypothesis. Hence either q-n = 0
and/or m-p = 0. If only either one is 0 then we get a or b is a zero vector
which, again, we know they are not by hypothesis. Hence the only possible
non-contradictory conclusion is that both q-n and m-p are zero. In this
case we get 0 = 0 which leads to the conclusion.
IMO the 3rd line is a bit misleading. One should add to the argument "assume
m != p and q != n" then ...." and the 3rd line arrives at a contradiction
because then we have a is parallel to b. Therefor the assumption that m != p
and q != n was wrong. and we must have m = p and/or q = n and it follows
almost immediately that if either one is true the other must be.
i.e., Suppose m = p then ma + nb = pa + qb ==> ma + nb = ma + qb ==> nb = qb
==> n = q. Similar for n = q.
My proof would go something like this:
"Let a and b be two non-zero non-parallel vectors. If ma + nb = pa + qb then
m = p and n = q.
Proof(by contradiction):
Assume m != p wlog,
ma + nb = pa + qb
ma - pa = qb - nb
(m-p)a = (q-n)b
then q = n ==> (m-p)a = 0 ==> a = 0. (which contradicts hypothesis)
Hence m = p.
QED"
one could take the other case n != q and arrive at b = 0 also. This is what
the wlog(without loss of generality) means since the argument would simply
be duplicated but with different symbols. (exact same logic and hence does
not need to be stated twice) Both implie the m = p(first one) and q = n(the
implicit one from the wlog part).
Similarly if we know m = p we get q - n for free(almost) since
ma + nb = pa + qb
==>
ma + nb = ma + qb
==>
nb = qb
==>
n = q
Gerry Myerson
Virgil <Vir...@home.esc> wrote:
> In article <U1jVm.62206$ze1....@news-server.bigpond.net.au>,
> Albert <albert.xt...@gmail.com> wrote:
>
> > Hello,
> >
> > I'm trying to teach and introduce myself to vectors. I'm reading the
> > following:
> >
> > _Linear combinations of non-parallel vectors_
> > If a and b are non-zero, non-parallel vectors, then
> >
> > ma + nb = pa + qb
> >
> > implies m = p and n = q
> >
> > An argument is as follows:
> > ma - pa = qb - nb
> > (therefore) (m - p)a = (q - n)b
> > (therefore) a = ((q - n) / (m - p)b or b = ((m - p) / (q - n))a
>
> Note that a = ((q - n) / (m - p) b makes no sense when m = p and
> similarly b = ((m - p) / (q - n))a makes no sense when q = n, since
> division by zero is not defined or allowed.
>
> But if either equation does make sense (by not dividing by zero) then
> you have two non-parallel vectors which must be parallel.
>
> Thus each of those necessarily senseless equations must have been
> constructed erroneously by dividing by zero.
So perhaps what OP is missing is that (non-zero) vectors x and y
are parallel if and only if each is a scalar multiple of the other.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
Albert
> Albert wrote:
>> <OP>
> <Jon Slaughter's explanation/proof>
I get it now :) Thanks to all who have posted to this subject
(especially to Jon Slaughter)