Reputation
3,083
Next privilege 5,000 Rep.
Approve tag wiki edits
Badges
11 28
Newest
 Announcer
Impact
~64k people reached

Sep
12
comment Negation of a proposition of the form “not(p) & q”
yes, I think this is correct
Sep
12
comment Negation of a proposition of the form “not(p) & q”
If that what you expressed in English is "p & not(q)" then it is wrong because you have to express "p | not(q)". I think you have actually expressed "p & not(q)", so it is wrong.
Sep
12
comment Negation of a proposition of the form “not(p) & q”
not( not(p) & q) <=> (p | not(q)) or better $$\neg(\neg p \land q) \Leftrightarrow ( p \lor \neg q) $$
Sep
12
comment Showing Surjectivitity of $f(x) = x^3$
but then you should mention that it is countinous if this is a premise for you conclusion
Sep
12
comment coin problem with two coins, inductive proof
If you want somebody to discuss your proof it would be easier if you number your equation/inequations. I think you made a mistake, you cannot get $pab-pa-b \ \neq \ mpa + b(na+p-1)$ from the former equation. it should be $pab-pa-b \ \neq \ npa + b(mp+p-1)$ Also you should make your naming of variables consistent In the first part you use $n$ for the multiplier of $a$ int he second part you use $m$. This makes it more difficult to read the proof.
Sep
10
comment Proof of special case of Fermat's Last Theorem
The title is really bad. It does not say anything about the mathematicl content of the post.
Sep
10
comment Proof of special case of Fermat's Last Theorem
+1 I read it after the 3rd edit and it looks very clear and right for me.
Sep
9
comment Are there any mathematics “problem websites” similar to Project Euler?
This contains programming puzzles and challenges. There I found a link to ponder this. The current puzzle is a math problem. and i needed programming to find a solution of the current problem. Pen and Paper problems can be found on (imo-official.org/problems.aspx). Books about algorithms, number theory or numerical mathematics may contain the type of problems you are looking for
Sep
8
comment seemingly simple question about polar coordinates
-1 this is you 5th question on this site, it's time to use LaTeX
Sep
6
comment Suppose $x$ is an odd function and let $h = f \circ g$. Is h always an odd function?
what is $x$ in the title?
Sep
5
comment Showing Surjectivitity of $f(x) = x^3$
that is not a sufficient condition, e.g. $f:x \mapsto e^x $ is strictly increasing and unbounded but not not surjective
Sep
5
comment Showing Surjectivitity of $f(x) = x^3$
that is not a sufficient condition, e.g. $f:x \mapsto \lfloor{x}\rfloor $ is not surjective
Sep
3
comment Encyclopedia of integers
this is the link to the wiki numbers: en.wikipedia.org/wiki/Category:Integers many of them contain the note "This article may contain excessive, poor, or irrelevant examples..."
Sep
3
comment Prove that $(2m+1)^2 - 4(2n+1)$ can never be a perfect square where m, n are integers
I changed the signs because it seemd to be wrong. But I am not sure about the meaning of the previous comments. IfF my changes were wrong so please undo them
Sep
3
comment Where am I going wrong with this Boolean simplification problem?
you must not write down the whole truth table but only one of the rows of the truth table where they differ. for which A, B does the truth table differ?
Aug
30
comment Probability problem from FELLER'S book
I fixed a tpe and changed thedownvote to an upvote
Aug
28
comment Clarify Cartesian Products and Binary Operations
from where do you have this definition?
Aug
28
comment Clarify Cartesian Products and Binary Operations
Your definition "A Cartesian Product is a function f:X x Y --> Z , where..." is not a meaningfull definition of the cartesian product because it is circular. You define the cartesian prodcut by using the cartesian productof X and Y
Aug
28
comment Explain rings and is [S, /, -] a ring?
You write "Binary Function" but link to "binary operation". Your definition of binary function if the definition of binary operator in wikipedia.
Aug
27
comment Simplify a boolean algebra expression: xy + xz' + x'yz
thx, I already understand. I did not look above enough.