i am relatively new to writing proof, i want to check whether this proof is good, and feedback on what can be improved on
#Show that $1(n-0) + 2(n-1) +...+(n-1)2 + (n-0)1 = \binom{n+2}{3}$
mild rose
delicate sierraBOT
i am relatively new to writing proof, i want to check whether this proof is good...
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pallid herald
i am relatively new to writing proof, i want to check whether this proof is good...
why is your background like that
mild rose
vscode
mild rose
lol
mild rose
yes a background looks good
pallid herald
- start by writing the LHS as a sum with sigma notation
mild rose
right yeah i did, but it didnt help with me that much
pallid herald
- *3 elements {a, b, c} as (a, b, c) might be taken as an ordered tuple
mild rose
$\sum_{k=0}^{n} (n-k)(k+1)$
upper treeBOT
pallid herald
best to let S={1,...,n+2} for ease of use too
mild rose
i see
pallid herald
now to the actual substance of the proof
how does picking n "b" 's divide S into 2 sets?
do you mean, "it leaves 2 elements"
mild rose
basically my logic was picking some middle number, a start number and an end number
say we have 1 2 3 4 5 6
if i pick 2 as b, then i have 1 and 3 4 5 6
so we have 1*4 ways to get the a and b
if i pick 3 as b, i have 1 2 and 4 5 6
so 6 ways
this was what i meant
pallid herald
so we have 1*4 ways to get the a and b
why?
we have 2 as b. We can pick a in 5 ways from 1,3,4,5,6
it'd be the same if b were 3
@mild rose
you there?
mild rose
you there?
i had to go to sleep
mild rose
we have 2 as b. We can pick a in 5 ways from 1,3,4,5,6
no, i pick 3 values
if we choose b=2 and from that divide the set into 2 sets, {1} and {3, 4, 5, 6}, we can pick a from the first set, A, and pick c from the second set, B
sorry for the cutoff
sudden badge
yes a background looks good
dope, I want that too, please teach me
The idea of the proof seems fairly solid
also, as coffey said, it is enough to pick S = {1, ..., n+2}
So that you can explicitly mention what A and B are
Namely, we can count the ways of selecting (a, b, c) from S = {1, ..., n+2} by:
- distinguishing cases over each choice b = 1, ..., n
- for each choice b, denote then denote A_b = {1, ..., b-1}, and B_b = {b+1, ..., n+2}, then ...
Though, it isn't entirely clear why you can only pick a from A and c from B
as in, why can't you pick them both from A
sudden badge
$\sum_{k=0}^{n} (n-k)(k+1)$
out of curiosity, can't you evaluate this directly?
$\sum_{k=0}^{n} (n-k)(k+1) = \left(\sum_{k=0}^{n} (n-k)\right) + \left(\sum_{k=0}^{n} (n-k)k \right)$
upper treeBOT
Rion
sudden badge
The first term easily n(n+1)/2
mild rose
dope, I want that too, please teach me
vscode background extension
sudden badge
the second term you can probably cheese too
mild rose
out of curiosity, can't you evaluate this directly?
no, i am supposed to reason that they are the same answer to counting questions
these are exercises on combinatorial proofs section
sudden badge
I see, however my question remains unanswered
regarding why you cannot pick both a and c from the same side
for a select b
or c from A and a from C, for that matter
mild rose
because it would get counted multiple times
sudden badge
basically you're enforcing a constraint that a < b < c
mild rose
yeah
sudden badge
Also, in terms of the notation |A|
ignore this if it is irrelevant, but the cardinal of a finite set, which is equal to the number of elements in that set, is actually not defined as that
as in, it is defined as something else but just happens to match
when a set A is in bijection with a set in the form {1, ..., n}, then the cardinal is defined as that only n
it's really weird shit, but basically since mathematicians decided that it wasn't easy to count in arbitrary sets, they would instead count by doing 1-1 pairs with all elements of {1, ..., n}
and then there's a bunch of elementary theorems about said mappings, blabla
mild rose
i see, but the book author seems to not mention that when he gave the definition of cardinality
sudden badge
anyway, taking S = {1, ..., n} will remove all the ambiguity about both cardinality and ordering
mild rose
though i understand why
sudden badge
though i understand why
Counting is a very difficult task, sometimes you don't even know what you're dealing with
but counting by doing bijective mappings to a set you can actually count, seems to work in our mind
the reasoning generalizes to countability in general, so a set is countable when it is in bijection with N
pallid herald
vscode background extension
ah that