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location Valdosta, ga
age 22
visits member for 2 years, 1 month
seen 10 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


Nov
22
comment Four Isosceles Trapezoids
Oh, wait! When I wrote the problem statement, I forgot to write that $x$ and $y$ are assumed to be positive integers. How does that change things? Could I immediately infer that $\displaystyle A = \frac{y+x}{2} \cdot \frac{y-x}{2}$ means that $A$ has to be factored into two positive integers?
Nov
17
comment Four Isosceles Trapezoids
I do not understand why I have to argue that $\frac{y+x}{x}$ and $\frac{y-x}{2}$ must be integers. I want them to be integers, so I compel them to be integers; and for each fraction to evaluate to an integer, the numerator must be some multiple of $2$. Is this the sort of proof you had in mind? If not, I do not see how it is possible to prove that they must be integers.
Nov
15
awarded  Notable Question
Oct
23
awarded  Popular Question
Oct
19
awarded  Notable Question
Oct
7
awarded  Notable Question
Oct
5
awarded  Yearling
Oct
1
awarded  Notable Question
Sep
24
awarded  Autobiographer
Sep
22
awarded  Taxonomist
Sep
15
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