207 reputation
19
bio website joelchristophel.com
location Reading, PA USA
age 19
visits member for 1 year, 2 months
seen 2 days ago

Nov
21
awarded  Yearling
Nov
21
comment Counting 5-letter words that include at least one vowel
While some of the other answers were very helpful, this solution most clearly demonstrates how the overcounting happens with the erroneous solution.
Nov
21
accepted Counting 5-letter words that include at least one vowel
Nov
21
comment Counting 5-letter words that include at least one vowel
My rationale behind the downvote was that the part of this response aimed at answering my question was incorrect. I already knew the correct method, which I cited in my question.
Nov
21
revised Counting 5-letter words that include at least one vowel
added 8 characters in body
Nov
21
asked Counting 5-letter words that include at least one vowel
Nov
10
awarded  Excavator
Nov
10
revised why is ${n+1\choose k} = {n\choose k} + {n\choose k-1}$?
th superscript, em dashses, removal of double space
Nov
10
suggested suggested edit on why is ${n+1\choose k} = {n\choose k} + {n\choose k-1}$?
Oct
25
awarded  Curious
Oct
24
asked Prove that $gcd(a, b) = gcd(b, a-b)$
Oct
22
awarded  Critic
Oct
22
asked How to prove a compound number is prime
Oct
8
comment Show that there exists a bijection from $(0,1)$ to $(0,1]$
Yeah, but $\frac{1}{n}\wedge2|n\wedge n\in\mathbb{N}\Leftrightarrow\frac{1}{2^{n}}:n\in\mathbb{N}$, silly
Oct
8
accepted Show that there exists a bijection from $(0,1)$ to $(0,1]$
Oct
8
comment Show that there exists a bijection from $(0,1)$ to $(0,1]$
Revision to the above: $2x$ if $x=\frac{1}{n}\wedge2|n\wedge n\in\mathbb{N}$. Will you confirm if I get it right?
Oct
8
comment Show that there exists a bijection from $(0,1)$ to $(0,1]$
Funny indeed. So I'm feeling the missing predicate is $x=\frac{1}{n}$ for $n\in\mathbb{N}$ because when $x = \frac{1}{2}$, $y$ should be $2(\frac{1}{2}) = 1$, etc.
Oct
8
comment Show that there exists a bijection from $(0,1)$ to $(0,1]$
Of course I wanted that. But it is fine that you did not provide that. I just thought you were attempting to provide it. Simple miscommunication :)
Oct
8
comment Show that there exists a bijection from $(0,1)$ to $(0,1]$
I understand your first sentence. But I don't understand "$f(x) = 2x$ for some values of $x$." Isn't "some values of $x$" ambiguous?
Oct
8
asked Show that there exists a bijection from $(0,1)$ to $(0,1]$