1,468 reputation
940
bio website
location Valdosta, ga
age 22
visits member for 1 year, 11 months
seen 14 hours ago

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


18h
accepted The role of 'arbitrary' in proofs
Sep
8
awarded  Popular Question
Sep
3
awarded  Popular Question
Jul
21
asked Restricting Binary Operator $*$ To A Subset
Jul
17
awarded  Notable Question
Jul
15
comment Proving the Well-Ordering Property
Why not? You just do the same comparison that I did in the case of two and three elements.
Jul
15
comment Proving the Well-Ordering Property
But I am proving that $S$ is well-ordered, and eventually, after having added enough elements to $S$, won't I get that $S = \mathbb{Z}^+$? So, because the two sets are equivalent, any property that one thing possess, the other thing must possess, right? In particular, $S$ possess the property that it has a minimal element; so, by the equality, $\mathbb{Z}^+$ has a minimal element.
Jul
15
comment Proving the Well-Ordering Property
Why isn't it clear what $S$ is supposed to be? Didn't I specify that it was some subset of $\mathbb{Z}^+$, and only contained the two elements $a,m \in \mathbb{Z}^+$? Isn't that clear enough? All I am doing in my proof is continually adding more elements from the set $\mathbb{Z}^+$ into $S$. How is that any less clear than what you said about $A$ in your post?
Jul
15
asked Proving the Well-Ordering Property
Jul
14
awarded  Popular Question
Jul
13
comment Proof About Division of Integers
That is what I figured, that the problem was concerned with existence, but I was not exactly certain.
Jul
13
asked Proof About Division of Integers
Jul
11
comment Inductive proof of inequality $a\le ab$ for nonnegative integers
I have another question, then. How does one prove that the sum of two positive integers yields a number greater than the individual numbers.
Jul
11
comment Inductive proof of inequality $a\le ab$ for nonnegative integers
Yes, we are dealing with $\mathbb{N}$. So, my reasoning is correct?
Jul
11
asked Inductive proof of inequality $a\le ab$ for nonnegative integers
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Jun
25
comment Considering Vectors Geometrically
You seem to be saying that the geometric picture follows from the way in which vector addition is defined, and the fact that $\mathbb{R}^n$ itself has a geometric interpretation?
Jun
25
asked Considering Vectors Geometrically
Jun
19
awarded  Popular Question