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12:08 AM
What did munchkin threaten this time?
hopefully no international incidents, we've been trying to keep those to a minimum
Wise …
apparently we're done with a new cat monster
It’s in beta production?
i really hope she isn't drawing these at school
when we draw a cat, she always draws a litter box. "in case she needs it." she also says "it can't see!" until i put pupils in the eyes.
kids are something else.
12:51 AM
hello, 2x4x8x16
No no
Given that we have $n$ jobs with start time $s_i$ and finish time $f_i$ for a job $i$ from $n$. Each job has a weight/benifit $b_i$. So the complexity here is $O(n2^n)$. I see that given a set, we have $2^n$ possibilities, but where did $n$ above came from :/ please
complexity of what?
alternatively, what's complexity?
i have to act like a mathematician. it's in my contract.
It's what the total number of possibilites :(
Very similar if not same concept
in this problem specifically
So, a list of $n$ elements has 2^n permutations, but still wonder what $n$ before $2^n$ there stands for
How is your daughter?
she's fine. keeps yelling about being hit by the cat. they do this every evening.
12:59 AM
cats can be evil sometimes
especially if they see a lot of math symbols around :{
they work together. they're plotting something.
Just keep Screech out of the plotting!
Most importantly they do math
Plot, symbols like algebra, ... those are good signs then
They are approaching calculus
1:30 AM
For this logical expression without truth table please:
$(𝑝 → 𝑞) → 𝑟$ is equivalent to $(𝑝 ∧ 𝑞) → 𝑟$ or not?
Answer based on truth table is NO. But any hint without truth table please?
Figure it out for yourself from the truth table.
Thanks Prof.
I did that. Just looking for what logical equivalences used above to conclude that is not the case
Let me try another one pls
Why is the answer no based on the truth table?
Answer is true b/c the combination of statements result is not the same
After you combine all trues and falses
It's not hard based on truth table. Just fill in values and combine
I was looking logical equivalences above, did not see how to infer that from it
I don’t see a truth table. Give me the scenario where they are not equivalent.
You’ve just copied from somewhere a bunch of equivalences.
1:41 AM
This is the truth table
They are not equivalent as I see it
Prof the above equivalences are there just to explain that I tried to match the question with them to see why they are not equivalent
But as you see truth table needs a lot of time to compute
So, this is why I am trying to figure it out from logical equivalences
Seeing that they’re different when all three are false is the right thing to do. Working with endless equivalencies seems hopeless unless you first have this insight.
You prefer to go with table pls?
Until you know what things to look for, how are you going to learn? There’s no replacing practice.
You seem to want shortcuts all the time and recipes for procedures. This is not how mathematicians proceed.
Prof I saw that from above we have $(p \cap q) \to r$ is equivalent to $p\to r \cup q\to r$
based on above equivalences, so I was trying to show why
(p→q)→r is not the case
So I made some progress :/
I do. Not believe this.
1:52 AM
Thanks anyway
Your first line there is true? Really?
Is $x^2=4$ and $x\ge 0$ Implies $x=2$ equivalent to $x^2=4$ implies $x=2$ or $x\ge 0$ implies $x=2$?
Maybe I really do not know logic and formal proofs.
Yeah prof
we have (p∩q)→r is equivalent to p→r∪q→r
this is true based on set of equivalences I uploaded before
23 mins ago, by Avra
user image
It's the last one from the bottom
I just gave one where both implications are false.
Yes. Thanks.
However, not sure if table above means those are the unique!
That's fine Prof
The idea is even your example needs time like table
It's recommended to do this in one minute :/
Not recommended but MUST
Well, I fail.
2:06 AM
Nope. Thanks. But this is our life :(
Good night Prof
I hope now you undestand me better :(
You might see that sometime I shortcut formulas, but this is the reason as you see above
In our case, you are not required to do solid proofs
You won't survive
Just get shorcuts and use them quickly
@TedShifrin. !!
I got something above
(p->q)->r and (p and q) ->r, what we could do quickly is to compare only first parts
I mean just match truth of first values if we have F->T->T, then in the second part it's F AND T -> T
maybe this is the best approach to solve it quickly
I mean we got just focus on first part of both (p->q) and (p and q)
2:50 AM
Hello I want to know the software that mathematician use to draw diagram for geometry please ping me
our professor require us to draw in computer for proofs
3:12 AM
@BannedUser What sort of geometry? Like high school geometry? Use Geometer’s Sketchpad.
@leslietownes but we don’t know if it’s right!
Please don’t please
3:29 AM
When we find limit of some function is it exact value of the slope of the secant at say x tends to zero
3:59 AM
Suppose $f$ is differentiable at $a$ and $g$ is differentiable at $f(a)$. Let $\epsilon$ be given. Then, there exists a $\delta_1$ such that $|h|<\delta_1 \implies \left|\frac{g(f(a)+h)-g(f(a))}{h} - g'(f(a))\right| <\epsilon$. Since $\lim \limits_{h \to 0} f(a+h) -f(a) = 0$, there is a $\delta_2$ s.t. $|h| < \delta_2 \implies |f(a+h)-f(a)|<\delta_1$.

Suppose $|h| < \delta_2$. Then, $f(a)+f(a+h)-f(a) = f(a+h)$, so $\left| \frac{g(f(a+h))-g(f(a))}{f(a+h)-f(a)} - g'(f(a))\right| < \epsilon$, and so $\lim \limits_{h \to 0} \frac{g(f(a+h))-g(f(a))}{f(a+h)-f(a)} = g'(f(a))$. Since $\lim \limits
does anyone believe my chain rule proof
the only problem with this proof is that $f$ may be constant around $a$.
I thought that could be a problem, but since $\lim \limits_{h\to 0} \frac{g(f(a+h))-g(f(a))}{f(a+h)-f(a)} = \lim \limits_{h \to 0} \frac{\frac{g(f(a+h))-g(f(a))}{h}}{\frac{f(a+h)-f(a)}{h}}$, we see the limit is equivalent to $\frac{(g \circ f)'(a)}{f'(a)}$, which is unproblematically defined, no?
oh, $f'(a)$ could be $0$
where the denominator makes sense only if it is non-zero
I don't remember the example...
it's in Spivak's chapter on derivatives.
you're right, since $f'(a)$ could be zero, the fraction is not always defined
being constant around a is not the only problem.
The problem is around $a$, for every $h\gt 0$, there may exist some $b_h$ in $(a,a+h)$ such that $f(b_h)=f(a)$
4:14 AM
thanks for the tips!
:) I like to prove Chain rule :)
Actually I am on my mobile . I can’t see your symbols. What do you want to say?
@shin: you may also wanna check out Caratheodory theorem.
I don't know why the link is not getting pasted.
4:59 AM
@Koro: the é needs to be escaped to %E9
it is illegal to do maths on your mobile.
happy to have my son home for the weekend.
cool. check the back door, he may not actually be home.
he actually conversed on the 2h ride from uc sc
he drives a little fast for my comfort (in our old pilot) but he is 18 with 3mo of experience.
hard to judge from him, but the ucsc academics don't sound up to much yet.
beautiful campus. lot of attractive moving scenery too.
not that i would notice.
the first few semesters of UC are pretty bad, in my experience. no effort put in.
on the instruction side. you're not going to win friends and influence people by kicking ass at teaching subject 101. or subject 1A, or whatever they call it
so you're in a room with a few hundred other people and everything kind of sucks. i think this is valuable life experience.
except the room is your door room and you are staring at your screen...
the only in person is a study group that my son set up.
5:14 AM
ugh, haven't even thought of that.
good for him.
hope the patient is doing well.
in grad school it was sometimes funny to see the yale and princeton kids, with their first TA assignments ever, adjust to the reality that not everybody in a UC freshman math class is a baby mathematician or cares one iota about anything.
she went to school without her stroller today. she still walks funny but this is apparently normal.
takes a while i am told.
on monday she goes back to full-time day care, which we had been avoiding when she needed a lot of help with toileting.
hip displasia is apparently a thing in our family gene pool. we have had quite a few casts.
5:16 AM
must have a german shepherd in the family.
i sidestepped the issue by matching genes from half way across the world.
our cat now sleeps in the daughter's bed for about half the night. she's there now. i can see her cat eyes glowing on the baby monitor.
i got some wine today, first time in months. trying to figure out what to pair it with.
met some friends for port & cigars last night, pleasant interlude.
i brought an Australian port as a backup which was good as the host's port better served the cooking community.
i am not good with wine pairings.
dry whites. probably going to pair them with a lentil soup or some kind of lentil curry over rice.
maybe a beef stew, if i can stop being lazy and cook one.
5:31 AM
Q: Does $\displaystyle\lim_{x\to 0} \frac{\sin(x^2\sin \frac{1}{x})}{x^2\sin \frac{1}{x}}$ exist?

Dr.  Jacob.Z.LeeThere are two definitions on the limit of a function: Definition 1: Let f(x) be a function defined on a neighbourhood of $x_0$, $\lim_{x\to x_0} f(x)=A $ if only if for every real $\varepsilon>0$, there exists a real $ \delta>0$ such that for all $\mathrm{x}, 0<|x-p|<\delta$ implies that ...

I have never seen such definition (definition 1 in the linked post) of limit.
Limit of a function is defined at a limit point of domain of $f$.
ugh, this is a well known thing. the interval-based definition is fairly standard, and pre-dates definitions where the domain is something more complicated than that. some of bourbaki's books use the goofy definition, i thnk. it fiddles with the answer to the questions like this.
it's not interesting enough for a consensus to have developed, i think people just state what they are familiar with, even in textbooks. pick the definition you like and run with it.
in this case, though we know that 0 is indeed the limit point of domain of f, the final answer should be without ambiguity only if the question asker says "there exists a $\delta\gt 0$ such that for all x in domain of S..." right?
Or there is no need to state that as it is understood (tacitly)?
you're putting your finger on the thing that is, or is not, present in definitions that changes the result
i don't know if it is or is not understood tacitly. i would file this under "be aware that it exists, but do not care"
it's never going to matter
this is my highly biased view, but, unless you're writing a 'history of definitions in analysis' textbook, you can safely avoid this, it's not going to come up
5:43 AM
another debated topic is: definition of discontinuity
whether it is to be defined on domain of function f or beyond also.
i think (personal opinion) that discontinuity should be talked about wherever f is defined and not beyond that.
just do the opposite of whatever bourbaki does
4 hours later…
9:26 AM
@Koro definition 1 is the standard definition of a limit. Definition 2 is the weird one
What is problem 2 known as?
2 hours later…
11:38 AM
how are we doing my dudes
3 hours later…
2:25 PM
2:55 PM
Hello, if we are to hash English letters, any idea please why there is 31 below In the hash function:

$$hash = 31 \times hash + c \in \{65, 122\}$$
where the hash is iinitially set to $hash=0$.
Answer found: The value 31 was chosen because it is an odd prime. If it were even and the multiplication overflowed, information would be lost, as multiplication by 2 is equivalent to shifting. The advantage of using a prime is less clear, but it is traditional. A nice property of 31 is that the multiplication can be replaced by a shift and a subtraction for better performance: 31 * i == (i << 5) - i. Modern VMs do this sort of optimization automatically.
Any though please?
3:57 PM
@robjohn Not so weird. Just limit with subspace topology. But not appropriate for beginners.
we all agree!
i think i'm mellowing out
Oh oh. Tell Munchkin to get busy.
she's drawing a cat monster
avra: looks like that gem of programming, the hard-coded constant chosen more or less at random.
1 hour later…
5:32 PM
@TedShifrin I was commenting on Koro calling definition 1 weird. definition 1 is more common, and if anything is to be called weird, it would be definition 2.
or as i referred to it, the 'goofy' definition
I wonder if the Donald would accept Goofy as a running mate.
definition 2 might not be weird, but it's goofy
Oh, @robjohn, serves me right for not reading everything. These people who learn a little analysis start to think they're gods ...
just prove it by contrapositive
5:46 PM
You mean by contradictory contrapositive.
5:57 PM
@robjohn Whoever is chosen as his running mate needs to know they'll be sacrificed in a heart beat! And Attorney General, and Cabinet Members, and, and, ....
@TedShifrin Ha?
@leslietownes She still wanna be a strawberry with claws?? For halloween?
6:31 PM
@amWhy I was chewed out on main by some students of formal proof for not understanding what a "correct" proof of contrapositive is.
If you're curious, here it is.
Hey, quick question. Is it also true that $\left(a+b\right)\pmod c = a\pmod c + b\pmod c$?
If you interpret the right-hand side correctly. Of course.
If your definition of mod is the CS definition of mod, then it won't quite work.
For example, $(4+2)\pmod 5$ is $1$, but what do you say $4\pmod 5 + 2\pmod 5$ is?
@TedShifrin TIL that every proof of $\lnot A$ is by contradiction...
@Lukas? TIL?
today I learned
6:37 PM
Today I Learned
Oh, thanks, folks :P TIL what TIL is.
New information
Zoomer acronym.
@TedShifrin I would still say it is 1, but if this is incorrect, then I am mistaken in my understanding of modulus here and what this is actually saying.
Well, my point is that you have to apply $\pmod m$ again after adding.
Cool, that's what I thought.
6:39 PM
To mathematicians, $k\pmod m$ is a whole infinite set of integers.
So the equation holds for those equivalence classes. And this is an important construct.
That makes sense.
@Ted I totally agree with you that the description of contraposition in the linked post is flawed
It would seem modulus has a distributive property?
At least in a notational sense
It obeys all the rules of algebra, @AMDG. That's the whole point. $\Bbb Z_m$ ($\Bbb Z$ mod $m$) is a commutative ring. Everything works. Addition, multiplication, additive inverses, etc. And when $m$ is prime, you get a field.
@Lukas What are these "formal logicians" doing?
I don't know
6:41 PM
Besides being condescending to me.
I like the compactness theorem from model theory
I love that.
but actually formal logic is not really my thing
Well, model theory is not "formal logic." It's substantive mathematics.
I think these people give logic students a bad name.
yes that's true
6:43 PM
I dropped the graduate logic class, having learned the compactness theorem and hyperreal numbers, after the third week on Turing machines to prove Gödel incompleteness. That was something like 1973.
I never went near a logic course again.
Have you heard of the LEAN project?
@TedShifrin So then if I understand correctly, if a set is a commutative ring, then all the usual rules of elementary arithmetic and operations apply to its members but with greater or fewer constraints (such as in this case, the modulus of the result of the addition)?
When you work with the correct definition ($k\pmod m$ as an equivalence class, NOT as a single number between $0$ and $m-1$) everything works correctly with no warnings.
This is very much worth understanding.
We say $a\equiv b\pmod m$ if $a-b$ is a multiple of $m$. Simple.
I think specializing to $m=2$ can be helpful to get an understanding of what happens. You have two equivalence classes: even and odd integers. All the rules in the ring of integers mod $2$ are familiar properties about parity: even+odd=odd, even*odd=even etc.
6:46 PM
Yes, thanks for that comment :)
I can't say I can view it merely as a number in $[0, m-1]$ from an intuitive position either. $k$ and some other value equivalent to $k$ in modulus of $m$ must be considered as distinct at least from a computational perspective. If I had to describe it informally, it is more so describing position relative to a multiple of $m$ just considering the graph of one of these rings or fields for a given modulus.
@amWhy yes, scary strawberry with claws.
That's why I distinguish between the math understanding of this and the CS understanding of this.
Of course, mathematicians too use the equivalence classes of $0$ to $m-1$ to enumerate the set.
it's helpful to be able to express modular arithmetic facts without performing the math equivalent of memory management, choosing representatives, etc. it really helps to have the language for this. how you implement something specific is a separate issue.
weird moment at the duck pond today. there was a dead heron in the water. prompted a conversation i was not expecting to have. she somehow seems to know a little bit about this already.
Ah, when does a child first conceptualize this notion? I have no idea.
6:58 PM
Yeah, I'm really starting to see the usefulness of modulus and how it ties to various things having looked for a good way to implement $2^x \bmod y$.
no idea. she kept asking how the heron 'got dead.' and then said that when a bird dies it should be thrown away 'like paper.' then she asked if dead herons were real.
all of this while we're trying to move on from standing right in front of one
Oh, is she at least walking now?
if you want to explain the math understanding in CS terms one might say that "$k\pmod{m}$" is not an integer, it's of a different type, namely of the type of "integers mod k". What a lot of programming languages do is to always directly convert "integers mod k" to integers which can be done by choosing a representative for each class. The most common choice for this is choosing $[0,1, \dots, m-1]$, but there are other choices as well
she has a weird limp that is supposed to correct itself, but yes, is walking.
Well, this is the dual discussion to the difficult one about when life begins.
6:59 PM
thankfully she already knows about storks. that'll be an easy one.
Yeah, but my only question now is how to make an efficient addition-subtraction-chain for computing this modulus of a power of two given that we can compute the first power of two for which $\frac{2^x}{y}\geq 1$.
Well, not so easy, actually. I didn't ask "where babies come from."
@AMDG Euler's theorem might be helpful if $m$ is odd
$y$ in this case
@LukasHeger I'm not looking for specific properties of $m$, just $m$ in general.
7:03 PM
I'm not sure that modular arithmetic is the right approach to this. It does not distinguish the notion of $k\ge \ell$ at all, since this is meaningless for equivalence classes.
While it is true that I could technically just keep doubling and computing the modulus up to $2^x$, that's a pretty terrible worst case for $2^{64}$ for any $y$.
Again, modulus tells you nothing.
@TedShifrin You are 100% spot on!! I answered this once. Or, many times on site; and probably 1000 times, at least, over my career.
I've learned of a manner to reduce this to a negative power of two that gives the equivalent of $2^x$ in $\pmod y$.
@amWhy Why are these pompous ***** messing up so many people? They are presumably undergraduates, but they think they understand the world.
This is a very confusing issue to beginning serious math students.
7:07 PM
Let's say you want to compute $2^{10000} \pmod{49}$. Now we have $\varphi(49)=42$, So Euler's theorem says that $2^{42}=1$. Now we just compute $10000 \pmod{42}$ and we get $4$, so we can conclude that $2^{10000} \equiv 2^4=16 \pmod{49}$
@leslietownes Can't forget that "scary" part! Mwhahaha
that's the method I meant
How does computing modulus tell me that $2^2 > 3$, @AMDG?
aspects of logic function like bait for a certain kind of person, where math is idealized as a space for correcting other people and catching them in errors.
I can now currently algorithmically determine, for example, that $2^{7}\bmod 11 = -2^3\bmod 11$.
7:08 PM
As I said above, I just hope these twits never become mathematics teachers.
what's weird about this is that math professors/teachers are sometimes unfairly stereotyped in this way, but it's far less common among them than it is among students.
I discouraged my students from doing proofs by unnecessary contradiction (where you actually assume $\lnot Q$ but then never use it) ... I made them rewrite when I could.
@TedShifrin Yes. I once suggested we write an abstract duplicate focusing only on the the material conditional; and one focusing on the difference between proof by contradiction, vs. proof by contrapositive!
@TedShifrin that's just teaching good style
@TedShifrin I would assume the condition is satisfied by $k\bmod m = k$.
7:10 PM
I do not understand @AMDG.
Well in what way are you asking me that? My intent isn't to use modulus to compute comparisons.
How does using modulus tell you the first time $2^k$ exceeds $y$?
Oh, no it doesn't. I'm just stating that I have a means to do that. ($2^{\lfloor\log_2(y)\rfloor + 1}$)
Oh, this whole discussion confused me. OK, @AMDG.
7:14 PM
Is there anyway to get off this "list," sir? @leslietownes
yesterday, by leslie townes
user, i saw the swipe at lawyers. you're on the list.
And that this is the first step in reducing $2^x$ to another $2^x\prime$ equivalence class which is easier to compute the modulus of.
@amWhy Did someone give that OP the canonical example of the proof that $\sqrt 2$ is irrational? That's my standard go-to for a contradiction proof that can be done no other way.
you can become a platinum tier investor in lesliecoin. this rare opportunity is not extended to many.
LOL @user178758
Bruh, this is awful. Is there a nicer notation for a prime of $2^x$?
7:15 PM
I've been on leslie's list for so long that I'm proud of it.
@TedShifrin Yes, and the OP was committed to prove "some other way".
What in the world is "a prime of $2^x$"?
@amWhy: OK, good time for me to tip my hat and say "good luck" and depart.
It would read as "Two to the x prime" like a variable.
Utter nonsense to me.
Although I suppose I could just use x prime
7:17 PM
Oh, $x'$ ...
I kept thinking prime numbers.
See you, @TedShifrin !
lol I was worried about that but didn't think it would be necessary to distinguish from prime numbers
LOL, @amWhy. It is lunchtime, but that was in response to that stubborn OP.
i might venture a "you can't prove a negative" just to see what that drums up.
or i would have in my younger, more annoying days.
If you cannot use a different letter, there's nothing wrong with $2^{x'}$.
7:18 PM
that's $2^x \ln 2$ by another name.
@leslietownes Be careful where you tread, Sir! ;P
Good time for me to go to lunch.
I would think using the prime notation is clearer for things like immediate results derived from the original value unless it becomes ambiguous given the context.
i like primes. i was being facetious.
7:19 PM
Yeah, I know :P
No one cares about the derivative of $2^x$. Owned.
i'd say i would 'go to lunch' except my lunch is right in front of me. mentally i'm always out to lunch.
"Local professor turned professor emeritus for continued mental decline; admits before courts that he's 'mentally always out to lunch'"
That would be professor demeritus
That^ and it usually means, " let me chew on it for a bit"
Seriously, though, these relationships between powers of two and non powers of two is quite enlightening. I never would have guessed.
At least for mod y=11, $-(2^{n-1} y - 2^n x) = 2^x\bmod y$ for $n = x - (\lfloor\log_2(y)\rfloor + 1)$.
And it would seem that $\left(x\bmod -y\right)\bmod y = x\bmod y$ as well.
7:31 PM
@AMDG How are you defining the mod function for negative modulus?
Often, $x\bmod y$ is defined non-positive if $y\lt0$
i.e. $5\bmod-3=-1$
I mean, it works on desmos... but I'm not sure how to define it. desmos.com/calculator/twjw40kcee
@AMDG This statement will be true since $\bmod y$ is applied last on each side.
Try $\left(x\bmod y\right)\bmod -y = x\bmod y$
that may not be equal
Oh, I had a few things missing there for mod y = 11 by the way. Should be $\left(-\left(2^{n-1} y - 2^n \left( 2^{\lfloor\log_2(y)\rfloor + 1}\bmod y \right)\right)\right)\bmod y = 2^x\bmod y$
@robjohn Doesn't look like they are.
It just seems to equal $x\bmod -y$.
yes, they are non-positive?
Pun intended
8:37 PM
a non-positive pun includes a neutral statement :P
9:09 PM
For vectors aggregation please, if we have $x1= [1,2,3]$ and $x_2=[4,5,6]$, what is meant by aggregating vectors here please?
maybe you tell us? that's not a standard term. are there examples?
9:24 PM
"aggregate" is sometimes used as a synonym for "sum" @Avra
Sum and aggregate are too simple. We must increase in sophistication. Sum = agglutination now. It must be pronounced with a British accent and you must be wearing a monocle and holding tea with outstretched pinkie. You must also be adept at market manipulation and end all of your playthroughs with "perfectly balanced game with no exploits".
I have just been informed from our overlords, IEEE, rulers of mathematics and computation, that the penalty for breaking this standard is death.
Of course, they're just rulers. I'm a protractor, so I have the upper hand with complex arithmetic.
Has anyone else managed to get a dialogue box from YouTube on a paused video tab that says, "Video has been paused. Continue playing?"
The funny thing about it is they forgot to add a No option. It just has the option for Yes.
you can always smash the laptop.
Or shutdown YouTube.
1 hour later…
10:41 PM
How is the weekend, @copper?
\o @RyanUnger
11:11 PM
@TedShifrin Awesome :-). My son drove me back from UCSC yesterday evening, we had dinner. Had a lovely ride in Tilden this morning before he got up and we had brunch together. More effective time together in two days than during a normal month! Should turn 61 every day...
Well, great. I’m so happy for you both!
Is he driving you back with him tomorrow?
How’s the Irish traveler doing?
11:32 PM
i have a very hastily and poorly tarmacked drive, i'd like to know where he is too.
11:47 PM
@TedShifrin He will be returned Monday, so I have him tonight. Tomorrow night some of his friends had requested that he cook for them, so I am happy for him to do that (plus it sort of incentivizes his return to Albany). Our Irish visitor Ultan (the term traveler carries quite the connotation for Irish folks) is doing very well, he managed to snag two nights camping in Yosemite, so he is getting & enjoying the full experience. Thanks for asking!
all that and a birthday. pretty cool.
it's birthday season around here, munchkin was last week, my wife and i are a few weeks from now.
nice 🎂🍴🎂🍴🎂🍴
i have a lot of friends who have birthdays around this time. my closest buddy growing up was born on exactly the same day.
me, my wife, and best friend from childhood have consecutive birthdays.
you have to get fairly deeply into astrology to tell us apart. some of the moons were in different places.

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