Mack

1,223 reputation

bio	website
	location	Saratoga, NY
	age	20
visits	member for	7 months
	seen	May 21 at 11:46
stats	profile views	203

I am just a man who has an insatiable desire for knowledge.

"For the rest, brethren, whatever is true, whatever is worthy of reverence and is honorable and seemly, whatever is just, whatever is pure, whatever is lovely and lovable, whatever is kind and winsome and gracious, if there is any virtue and excellence, if there is anything worthy of praise, think on and weigh and take account of these things [fix your minds on them]."~ Philippians 4:8

	1,223 reputation	bio	website		visits	member for	7 months
7 badges		location	Saratoga, NY		seen	May 21 at 11:46

summary answers questions tags badges favorites bounties reputation activity

264 Comments

suggestions reviews revisions comments badges posts accepts all

Apr 25	comment	Solve for $x$: $4x = 6~(\mod 5)$ @AyushKhaitan What precisely is an integral value?
Apr 25	comment	Solve for $x$: $4x = 6~(\mod 5)$ It is? How so?.
Apr 25	comment	Solve for $x$: $4x = 6~(\mod 5)$ I don't see that written anywhere in your answer.
Apr 25	comment	Solve for $x$: $4x = 6~(\mod 5)$ I'm not quite certain of what you are doing. Are you solving for a specific x-value?
Apr 23	comment	Using The Pigeon-Hole Principle So there difference being less than n--that is, $j - k < n$--proves that when taking the modulus the result is distinct? I'm sorry, I don't quite see it.
Apr 21	comment	Probability Distribution Of A Linear Combination Thank you, that certainly helps for the most part; but how do I find the probability distribution $X+Y$?
Apr 20	comment	Probability Distribution Of A Linear Combination Thank you for the suggest, @jay-sun. I will be edited momentarily.
Apr 20	comment	Polar Coordinates: Dividing by the variable “r.” So, since we integrate $dr$ from 0 to $2\cos\theta$, we don't have to worry about $2\cos\theta$ having to assume the value $r=0$, because the lower limit takes care of that? If this is correct, how can one explain this is fewer indefinite words, that is, explain it more mathematical. Also, what exactly do you mean by making a limit argument?
Mar 17	comment	Proving By Subsets Hmm, okay I see. I just don't understand why my teacher said that you can't use laws when doing a proof by subsets. And even after she said that, she did an example proof by subset using laws! I understand, now. Thank you, Ross, for your help.
Mar 17	comment	Proving By Subsets Okay, but if I was to strictly do a proof by subsets, I would just suppose $x \in~thing_1$, and describe how $x$ being in $thing_1$ also describes $x$ being in $thing_2$, and not use any sort of laws?
Mar 17	comment	Proving By Subsets Okay, I see. That's pretty much what I understood the process to be. Just to be clear, I am allowed to use laws (such as De Morgan's) while proving either side is a subset of the other?
Mar 17	comment	Proving By Subsets @lamb_da_calculus All right, that's pretty much what I understood the process to be. Just to be clear, I am allowed to use laws (such as De Morgan's) while proving either side is a subset of the other?
Mar 17	comment	Proving By Subsets I'm sorry, I didn't mean proof using subsets, I meant to say proof by subsets. Here is an example problem: Show that if $A$, $B$, and $C$ are sets, then $(A∩B∩C)^c= A^c∪B^c∪C^c$. And I am asked to prove that these are subsets of each other, thus proving that they are equivalent.
Mar 17	comment	Proving By Subsets I think it rather clear what I am inquiring into: I am asking about the general approach to proofs by subsets. Additionally, I thought that this would be better than posting a problem I am working on and asking help with it, without having done any work myself. The intimation towards her incompetency explains why she contradicted herself in class--and on several occasions, if I might add.
Mar 16	comment	Prove that $\mathcal{P}(A)⊆ \mathcal{P}(B)$ if and only if $A⊆B$. Oh, I see. Because $X$ is an element of the powerset of $A$, it must be a subset of $A$, etc. Thank you both for the help.
Mar 16	comment	Prove that $\mathcal{P}(A)⊆ \mathcal{P}(B)$ if and only if $A⊆B$. @BrianM.Scott For your first comment, why is $X$ a subset of both $A$ and $B$?
Mar 16	comment	Prove that $\mathcal{P}(A)⊆ \mathcal{P}(B)$ if and only if $A⊆B$. Whoops, I forgot to put these \ around my brackets
Mar 15	comment	Finding A c Value, Such That Vectors Satisfy A Certain Condition Oh, of course. I am terribly sorry for bothering you with something so trivial. Thank you very much for your help, @Muphrid
Mar 15	comment	Finding A c Value, Such That Vectors Satisfy A Certain Condition I guess what is confusing me is, why do we have to dot product of the vector and $\hat{k}$?
Mar 15	comment	Finding A c Value, Such That Vectors Satisfy A Certain Condition No, I believe I understand these notions. I just don't quite understand how procedure gives me a c value such the vector is perpendicular to the z-axis. In short, how do I know that this particular c-value is the c-value we need. I can do a lot of things: could probably set it equal to 5, and then solve for the c-value.