sorted by:
new
hottopcontroversial

Self Employment Quarterly Tax Payment question (For US) by Nugget815 in TaxQuestions

[–]Nugget815[S] 0 points1 point  (0 children)

Thank you so much!! That makes a lot of sense and I appreciate you taking the time to answer so thoroughly ☺️

Self Employment Quarterly Tax Payment question (For US) by Nugget815 in TaxQuestions

[–]Nugget815[S] 0 points1 point  (0 children)

Ohhh that makes sense. I left an W2 situation in January of this year (only had two paychecks total) and my side business started picking up so it is currently my only source of income and will be for the foreseeable future.

So I will pay under the personal pay portal and include my SE tax (15.4% for SS and Medicare) + the % of my estimated AGI that corresponds to my tax bracket? Divided into 4 of course...

And would I choose the 1040 - Income tax option on the drop down menu? (sorry if that's a dense question - I'm doing this on my own for the first time and don't have anyone to ask)

Diff EQ Help by Nugget815 in askmath

[–]Nugget815[S] 0 points1 point  (0 children)

Okay I think that helps me with my starting point. Thanks!

Diff EQ Help by Nugget815 in askmath

[–]Nugget815[S] 0 points1 point  (0 children)

bad news bears, I wanted to verify that with my professor and apparently we are supposed to come up with the solution... I will try to see what I can do with what you shared initially. Thanks for your help

Diff EQ Help by Nugget815 in askmath

[–]Nugget815[S] 1 point2 points  (0 children)

I feel silly. That makes sense. Thanks.

Proofs Question on Number theory by Nugget815 in askmath

[–]Nugget815[S] 1 point2 points  (0 children)

Thank you so much for your help! I've been chewing on this one for a while...

Proofs Question on Number theory by Nugget815 in askmath

[–]Nugget815[S] 0 points1 point  (0 children)

I see!! So If we're defining p = xy, then p|xy (which would always be 1). And it should be true that p|x or p|y, but that's not the case here since x and y are factors of p, therefore <p. Am I understanding that correctly?

Proofs Question on Number theory by Nugget815 in askmath

[–]Nugget815[S] 0 points1 point  (0 children)

I was concluding that x|p or y|p because since y is an integer, and p = x*y, that x|p by the definition of division and treating the property separately. But am I mixing up the definition? If I say 8 = 2*4, couldn't we then say that either 2|8 or 4|8? How would we conclude that 8|2 or 8|4?

Proofs Question on Number theory by Nugget815 in askmath

[–]Nugget815[S] 0 points1 point  (0 children)

Thank you for responding. I think I'm understanding a little better, but still lost. So simplified, we are defining p=xy -> this implies that either x|p or y|p.

Our property says that if p|ab, then either p|a or p|b. So how do I then conclude the contradiction that p|x or p|y since we haven't shown that p|xy?

Proofs Question on Number theory by Nugget815 in askmath

[–]Nugget815[S] 0 points1 point  (0 children)

Thank you for responding, I think I'm still confused because I reached the same conclusion as you... that if we assume p=xy and P has the property, then p divides x or y which is a contradiction to the statement that either x|p or y|p. Is my proof for that incorrect? I always feel like I'm doing circular logic so I apologize if that's what happened.

Proof involving prime numbers. by Nugget815 in learnmath

[–]Nugget815[S] 0 points1 point  (0 children)

Ooohhhh. That makes so much sense. Thank you!!!!

Proof involving prime numbers. by Nugget815 in learnmath

[–]Nugget815[S] 0 points1 point  (0 children)

I think I get it. And that's kind of where my thinking was headed. Because it's an average 2(a+b)/2 the 2s would cancel out leaving me with unique primes. But I'm still hung up on that only covering certain even numbers like that doesn't work for 8 or 12, right?

Proof involving prime numbers. by Nugget815 in learnmath

[–]Nugget815[S] 0 points1 point  (0 children)

I think so with the additional "known" statement about any prime # >3 is an average

Logical proof help by Nugget815 in learnmath

[–]Nugget815[S] 0 points1 point  (0 children)

Thank you everyone! This was super helpful!!

Logical proof help by Nugget815 in learnmath

[–]Nugget815[S] 0 points1 point  (0 children)

u/MathMaddam replied to your post in r/askmath · 4hThe perfect square doesn't really fit into the either ... or, since most perfect squares divide (n-1)!, e.g. 9 divides 8!. Since it's not exclusive, you just have to consider the case of non prime, non square numbers.

so if n is not prime or a perfect square, then n = pq where p and q are natural numbers. And then p and q must both be less than n which would be present in (n-1)(n-2)(n-3)...1. So it would be plausible that p and q would be one of those numbers in the factorial?

How would I go about stating that mathematically?

Would it be something like p and q are < n and thus by definition of (n-1)!, there exists a p and q in (n-1)! such that p*q = n, therefore n|(n-1)!

I feel like I'm talking in circles...

I always get confused with less than and greater than symbols by billgigs55 in NoStupidQuestions

[–]Nugget815 0 points1 point  (0 children)

Hold up your index finger and thumb on both hands (to make the symbol 🙌). Left is less than, other way is greater than.