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Unfortunately I have read few proofs and have written fewer. So I am asking for advice on how to write this one so hopefully I can get better at it.
I am to show that
Eq1: Abs[ D[y,x] ] + Abs[ y ] + 1 = 0
has no solutions.
So this is what I was thinking.
Let
f(y) = Abs[ y ]
g(y) = Abs[ D[y,x] ]
Thus we can rewrite Eq1 as
Eq2: f(y) + g(y) + 1 = 0
However due to the Abs operator the domain of both f and g
must be greater than or equal to zero therefore there is no y
in the range of f or g that can satisfy Eq2.
I guess first off, is this proof even correct?
Is there anything I should do to improve it?
Is ok for me to imply and assume the fact that 1 plus a
non-negative cannot be less than 1 thus the sum cannot be
zero? I suppose the answer depends on the audience. Here I
assumed the audience was my college student peers with
the material we have covered to date, including previous
classes, therefore the proof of this fact should be pretty easily
available.
I guess I am most concerned with where to draw the "it's trivial
from here" line. I mean, to be honest, this proof in itself seems
a bit trivial which, ironically, is what makes it hard to write. :)
I am to show that
Eq1: Abs[ D[y,x] ] + Abs[ y ] + 1 = 0
has no solutions.
So this is what I was thinking.
Let
f(y) = Abs[ y ]
g(y) = Abs[ D[y,x] ]
Thus we can rewrite Eq1 as
Eq2: f(y) + g(y) + 1 = 0
However due to the Abs operator the domain of both f and g
must be greater than or equal to zero therefore there is no y
in the range of f or g that can satisfy Eq2.
I guess first off, is this proof even correct?
Is there anything I should do to improve it?
Is ok for me to imply and assume the fact that 1 plus a
non-negative cannot be less than 1 thus the sum cannot be
zero? I suppose the answer depends on the audience. Here I
assumed the audience was my college student peers with
the material we have covered to date, including previous
classes, therefore the proof of this fact should be pretty easily
available.
I guess I am most concerned with where to draw the "it's trivial
from here" line. I mean, to be honest, this proof in itself seems
a bit trivial which, ironically, is what makes it hard to write. :)