- #1
musickps
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I am not sure if this is the right thread, since I am new. Bump it if necessary.
I am currently writing an essay on revising the conventional A-series of time to account for relativity and simultaneity, among other problems. In the course of my argument, I have used an algebraic statement. I know that the organization is not up to convention, but this is a rough draft. I am just trying to verify that this is a true statement. I seem to remember some transitive property from my high school days that suggests that it is, but I want to make sure before my supervisor sees it! Thanks for the help!
Assuming
(1) all given variables represent non-zero integers and
(2) no integer in the addition operation appears more than once
-Let s represent the sum of following integers: m, n, x, p
If (s)(z) = y
then the following expressions must all be true:
(m)(z) ≠ y,
(n)(z) ≠y
(x)(z) ≠y.
(p)(z) ≠ y.
Mods: I was not sure where to post this since it is algebraic in nature but I am using it in a theoretical manuscript. Feel free to bump it wherever it belongs. Thanks for the help!
I am currently writing an essay on revising the conventional A-series of time to account for relativity and simultaneity, among other problems. In the course of my argument, I have used an algebraic statement. I know that the organization is not up to convention, but this is a rough draft. I am just trying to verify that this is a true statement. I seem to remember some transitive property from my high school days that suggests that it is, but I want to make sure before my supervisor sees it! Thanks for the help!
Assuming
(1) all given variables represent non-zero integers and
(2) no integer in the addition operation appears more than once
-Let s represent the sum of following integers: m, n, x, p
If (s)(z) = y
then the following expressions must all be true:
(m)(z) ≠ y,
(n)(z) ≠y
(x)(z) ≠y.
(p)(z) ≠ y.
Mods: I was not sure where to post this since it is algebraic in nature but I am using it in a theoretical manuscript. Feel free to bump it wherever it belongs. Thanks for the help!