Reputation
1,050
Top tag
Next privilege 2,000 Rep.
Edit questions and answers
Badges
1 10 29
Impact
~75k people reached

Apr
27
comment Can this puzzle be solved without brute force?
in other words, it is possible for the product of two numbers to be an integer multiple of $ab$ without either number being an integer multiple of $ab$.
Apr
27
comment Can this puzzle be solved without brute force?
in your last step in your list of equations, you assume that if $xy = 0 \mod ab$ then either $x = 0 \mod ab$ or $y = 0 \mod ab$, however consider this: $2\cdot 3 = 0 \mod 6$, but neither 2 nor 3 is 0 mod 6.
Apr
26
comment Can this puzzle be solved without brute force?
@Dhruv Another solution for $(a,b)$ is $(6,3)$ because $7/3 + 4/6$ is an integer. To answer your first question, these fractions aren't integers though their sum is.
Apr
26
comment Can this puzzle be solved without brute force?
@Dhruv yes I meant "fractional parts." Thanks. I think you meant $(a+1)/b$ and $(b+1)/a$ though, because $a+1/b$ and $b+1/a$ can't be integers if $a$ and $b$ are integers.
Apr
26
revised Can this puzzle be solved without brute force?
added 6 characters in body
Apr
26
asked Can this puzzle be solved without brute force?
Apr
25
revised Antisymmetric Relations
added 20 characters in body
Apr
5
comment Help with proving a logical equivalence
@Anonymous because, as I'm sure you know, the first answer is not always perfect. it takes many revisions to get it right. I'll admit I took that one step from your answer, which at the time I wasn't sure of, but the important part is that I went back and proved it later on when I had time. I never take material I don't understand and claim it as my own. but if mathematicians weren't allowed to borrow material from one another, we'd still be in the Stone Age. get off your high horse and stop fretting over reputation points.
Apr
5
comment Help with proving a logical equivalence
@Anonymous as hard as it may be to realize that there are other people with intelligence in this world, I arrived at my solution independently from yours. I can send you screenshots of my hand-written scratch work if you'd like.
Apr
4
revised Help with proving a logical equivalence
added 163 characters in body
Apr
3
revised Help with proving a logical equivalence
deleted 93 characters in body
Apr
3
revised Help with proving a logical equivalence
deleted 93 characters in body
Apr
3
revised Help with proving a logical equivalence
deleted 123 characters in body
Apr
3
revised Help with proving a logical equivalence
deleted 123 characters in body
Apr
2
answered Help with proving a logical equivalence
Apr
2
comment Help with proving a logical equivalence
What have you tried first?
Apr
2
comment Help with proving a logical equivalence
This isn't an answer, this is a hint. It's a good hint, but still just a hint.
Mar
30
awarded  Notable Question
Mar
13
comment Write the negation of the following logical expression: ∀𝑥,𝑦∈ℝ,∃𝑛∈ ℕ,(𝑥−𝑦)𝑛<0
Your guess is correct. On the Stack Exchange network, you may provide an answer to your own question and perhaps even get it upvoted.
Mar
1
comment Prove $i\notin \mathbb R $
I found a simpler method. Assume $a < b$ and multiply throughout by $i$. If $i > 0$, then $ai < bi$, and do it again and get $-a < -b$, a contradiction. If $i < 0$ though, you get $ai > bi$, and again, $-a < -b$, the same result.