#help-23
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If you have any feedback I'm very willing to hear though 👌
Yeah grading my classmates work I've seen some horrors 💀 😅
Proofs with no words
All symbols 💀
i see both that and proofs with only words
I'm sure you would love some of my proofs from freshman year where I get distracted in the middle and have to come back later to finish
No no that makes sense, I've never heard of structural induction tbh so I was just making up what I thought might be right
@cloud hound Has your question been resolved?
Oh but thank you! I should say
I'll be continuing to do exercises from the book for at least a week
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x=10 mod 21
x=10 mod 22
x=10 mod 56
x=0mod37
x can be replace by 37k
looking for any kind of short approach which takes very low time
Find the least common multiple of 21,22,56 and express x in terms of the LCM
Then plug that expression into mod 37
LCM of 21,22,56 is 2^3.3.7.11
i did not understand express x in terms of LCM
you meant like 37
big divide wait
1848/37 so we get
Then just find lcm of those two numbers
i got 1848=37.49+35
How's that useful?
Do yk how to find lcm?
35=-2 mod 37
1848 is LCM
Now find lcm of 1848 and 37
please check my work
Check what exactly
x-10 is a multiple of LCM(21, 22, 56) and x-10 ≡ -10 (mod 37)
so we should be looking for the least value of x-10 in light of this, i think
@silk cove
x-10 ≡ -10 (mod 37)
ohh 1848k+10 mod 37
so we need to check by putting k=1,2,3,4
so on
k=5 works
ohh simple
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Question and book solution
I added both amounts then applied discounts. Seems right as per information given in question.
Am I wrong?
you're applying both discounts to both bills at once
in reality the 9800 bill is supposed to ONLY get the 2% off for 'within ten days' and the 5480 gets ONLY the 5% off for 'payment the day-of'.
these discounts don't stack in any way!
@strong pike Has your question been resolved?
Got it. Took me a while to understand English lol. Thank you.
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Can someone please walk me throught his dry run of the Dijkstra's algorithm
I mean I know how it works but I do not know how to fill this table
the table is for this graph
@midnight sonnet Has your question been resolved?
mhm, I'll try to explain it in the best of how I can word it. So you know the Dijkstra algorithm. It starts from a permanent point and adds a certain weight to each other point. V(T) is all the points you've crossed at step 1,2,... E(T) is the connection. F is a sort of buffer I believe, to see which points you haven't yet used. and L is the weight of each point at every step.
So at step 0 you know know that point a is 0 and all others haven't been checked so therefore they are inf
once you set that first step and check from a-b you know the weight of b
does this help? or is something in my explenation confusing?
@midnight sonnet
Thank you, I will read your explanation with the graph side-by-side, if I have any queries then I'll ask again
Alright, feel free to ping me when you're stuck
@livid lodge This is the graph and here is the table from before, now as you can see that after we assess the distances of b and c from a, we put F as {b,c} in the table. Which are the vertices that we have checked the distance of a from. But in step #2 of this table, we are checking the distances of d and e from b. Then why do we have to include c in F too? As you can see, F in step #2 is {c,d,e}, whereas I think that it should have been {d,e}, because we are only checking b's distance from d and e, but not c.
F is a sort of buffer, it stores the points not checked. So in Step 0 you have a in F, so that's the on you check in that step and fill in all the neighbours of a. These are 'temporary points'(idk how your professor explained Dijkstra, that's how he explained it to me).
So next step, you have {b,c} in your buffer, so you see the weight of each point from a. for b that's 3, c that's 4. Next step you choose the one with the lowest weight, in this case b(since b<c obv.) so we remove b from the buffer F and make it a 'permenant point'(meaning this is decided to be the shortest path from your origin point, in this case a). Then you repeat stap 0, check all the neighbours of b: d & e and fill their weight in. Once you've done that you now have the buffer F that started like this: a[origin, therefore smallest point] -> b,c [neighbours of a, with b smallers, take this one out of the buffer] -> c,d,e[neighbours of b, with c the smallest, take this one out].
This is a bit of extra info for the next step if you're still confused, text above should have everything already explained. So now we have c, we make that point permanent, and check it's neighbours, that's a and e. a has already been checked so doesn't get added to the buffer, e is already in the buffer so nothing changes. Check the weights of c to e, that's 1. Add the original weight from c + {{e,c}} and you get 5, this is FAR shorter than from a-b-e so it gets changed in the table. Now e has become the smallest point and rinse and repeat
If something is still confusing you, I'll try to visualise it more in the next explenation :) feel free to still ping me
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@midnight sonnet Has your question been resolved?
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"Hi everyone! I'm from the Philippines and I'm aiming to win the Division Math Celebration Quiz Bee for senior high school. I'm in Grade 11. I think the competition will focus on statistics what i would do
Please don't occupy multiple help channels.
what topics are you looking for?
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$\int^1_{-2} \frac{x+1}{\sqrt{x+3}+2}dx$
Professor Edward C. Hawthorne
How could one understand that the best substitution was √(x+3)=t?
I tried to rationalize it since there was a root underneath, by parts...
Also it's not clear to me after the substitutions, how you break the numerator and denominator like that
,rotate
,rotate
no idea about why he thought about taking root x+3 as t tho
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please give me a proof question
(either direct or contraposition or contradiction)
prove √3 is irrational
okies
if a^2 = 3n, then does a = 3m for some m?
of course it does, a = 3(n/a)
remember that each number has a unique prime factorisation
what if n/a is not an integer
does that break anything
then i suppose m wouldn't be an integer either, but what does that mean?
i define a multiple of 3 as any integer n which can be written as n = 3k for any integer k
does k here has to strictly be an integer?
well given that you defined it that way.... yes
well then n/a could not be an integer, then i cant say that a is a multiple of 3 if a^2 is
why can n/a not be an integer?
i'm having trouble following your train of logic; could you state it in a more self-contained manner?
that is my question too
will it always be an integer
if a^2 = 3n
why is n here
alright i'll just state the condition that i faced in the proof
this was the question
yea so squaring would give you this instead then
now i wanna conclude a is a multiple of 3
prove that 3 | n^2 —> 3 | n
(where a | b means that b is divisible by a)
a | b means b = ak?
yes
n^2 = 3k
n = 3(k/n)
n = 3l for l = k/n
btw in b = ak, is k always gonna be an integer
yes
so when i do k/n = l, l might not be an integer
how do you know that k/n is an integer
you wanted a proof by contradiction / contraposition question right?
you can in fact, use one of those to prove what knief is asking
so why don't you give it a go first and forget about hayley's question for now
then use this result to come back to it
why?
alright well contraposition proof works
premise: 3 not | n
then n != 3k
n^2 != 9k^2
n^2 != 3(3k^2)
n^2 != 3l
hence n^2 not | 3
so 3 | n^2 means 3 | n
yeah, the main idea being that is 3 does not divide n, then it cannot divide n^2 either
so could you perhaps use this now to prove hayley's problem from before?
do i have to include this in that proof
or is it implied (was very non obvious to me )
that depends on how much detail you want to put in, but it is worth knowing if it's non obvious to you
prove there are infinitely many primes
can i express the primes as any n belong to N such that: for all m belong to N, GCD(m,n) = n
if by N you mean naturals, then it is almost right; the only difference is that 1 is not prime
no
you should just think of primes as positive integers that are not 1 that are only divisible by 1 and themselves
express the primes as any n belong to N such that: for all m belong to N, GCD(m,n) = n why is this incorrect?
suppose you have say 3, an obvious prime
then gcd(5,3) = 1, not 3
can i express the primes as any n belong to N such that n | n and n | 1
wat now?
what
^
i did that
dont think about it like this at all
then how
this is enough, but also knowing that every number uniquely factors as a product of primes
okay how about we start like this
look at the question, do you think you would use contradiction or contraposition?
and why
contradiction, because there is no conditional here
there are a finite number of primes
okay, good
so one thing we can do with this, is suppose that we have this finite list of primes p_1, ..., p_n
if we can find a prime NOT in this list, then we are finished
we're trying to look for a way to achieve this
can you think of any methods
with this as a hint
well if for every n belong to N, there is a unique combination of product of primes, then we need an infinite amount of combination of primes to accomodate every n. That would mean that there are infinite primes
not quite, because we can make infinitely many numbers with even one prime! since we can just take 2^n for any n
what i am rather hinting towards is trying to find a number which is not divisible by any of these primes
how can we do that?
like suppose if 3 is the only prime
well i think we need infinite combinations of primes to accomodate all n
thats why there are infinite primes
right but that is not a proof
often times in math there are things that work very unintuively
so we can't just say "i think this will happen" and then call it a proof
isnt it true that we need infinite combinations
well can you prove that
its one of those conjectures or smth
if u have infinite numbers then u got infinite combinations
suppose our only prime was 2
then 2, 2^2, 2^3, 2^4, ..., 2^n, ... are only "combinations" of one prime
and yet you have infinitely many numbers
so this is not a valid argument for saying there are infinitely many primes
alright then how abt i add 1 in your list and then prove that it is prime
that p_1...p_n , i add all these, and add a 1
that does not prove anything
okay how abt i multiply them and then add 1
okay, why?
if i can prove that the new number is also prime then there is a prime that is bigger than my original existing largest prime
i.e a contradiction
okay, but what if it is NOT a prime
for example, 2*7+1 = 15 is not prime
but it still proves existence of a new prime... why?
if it is composite then there must be a prime from my existing list that divides the large number formed
but is every composite divisible by some prime?
that is precisely the definition of a composite number, yes
so if the product + 1 is composite, why does a prime from our list not divide it?
because there is a remainder 1/p
sure, we can say that
ok so the list is not complete
more precisely it's because it would have to divide 1 also, and there is no such prime
there is a prime that is out of our list that divides our number
exactly
yep :) nice job
no
here's the thing
how do i come up with all this rocket science in the exam hall
there is less time and more pressure and problems harder than this
ok... poor job then mb
okay well you know when to use contrapositive and contradiction
you know what assumption to make and you just think about why it's useful
i think, these are kind of natural proofs that you were asked so far
btw why are all composites divisible by some prime
what is a composite number
a number that divides more numbers than just 1 and itself
okay, and what if we keep splitting up this composite number into its divisors
well you want me to say that we end up with a prime
but idk why is that the answer to this question
our divisors keep getting smaller and smaller
yup
so eventually, we need to find a number which has no divisors other than 1 and itself
which is precisely what a prime is
alright
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a)
Since $a_n = \frac{a_{n-1}}{2+a_{n-1}}$ then $a_{n+1} = \frac{a_n}{2+a_n}$
$\frac{a_{n+1}}{a_n} < 1$ as $a_n >0$ it simply implies that it would be a decreasing sequence. We also know that $a_n > 0$ Hence, it is bounded by 0.
So, by the monotone convergent theorem, ${a_n}$ is convergent.
b) Since it's a decreasing monotone function, I inferred that $a_1 > a_2 > a_3 ... > a_{n-1} > a_n$. Since $\lim_{n\to \infty} [a_n] = L = \lim_{n\to \infty} [a_{n-1}]$ then
$\lim_{n\to \infty} [a_n] = \lim_{n\to \infty} \frac{a_{n-1}}{2 + a_{n-1}}$ is same as saying L = $\frac{L}{2+L}$? $\therefore$ the limit of ${a_n}$ as n tends to $\infty$ is 0.
Maddie
I tried this question and I would like to get inputs on whether I did it correctly or not?
The answer is 0 - textbook solution.
I can't say anything on correctness but I am wondering how you derived a_n+1 = in a)
So my thinking was if the $a_n = \frac{a_{n-1}}{2+a_{n-1}}$ it would mean that at some n, the value of $a_n$ is that of the previous value. Hence, $a_{n+1}$ would be $a_n$.
Maddie
I think I worded this really really badly 
Yeah
Its correct
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One noge
Note
You should explicitly mention why you chose to ignore -1
Otherwise, good job
Sorry, what do you mean by -1?
Oh I see it's not multiplication it's separate sentences
I thought it was multiplication 😅
Solving L = L/(2+L) gets you two solutions
Oh!!! Yes yes, thank you so much for your help!
you're welcome
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.solved
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Can someone check if this proof is right?
Ok thx
Waes (Wires)
Also $\in$
ExpertEsquieESQUIE
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is this a PHP question
What is PHP?
pigeon hole principle
idk what the buckets would be tho
maybe you could try to contrive a scenario where NOBODY is surrounded by two boys
I suppose yes, since the only way to avoid that would be to have girl-girl-boy-boy-girl-girl-...
And eventually the last spot would cause 2 boy neighbours
how to do that
will that invlove PHP
Like this
This
seats 23 and 2 will have to be girls bc then seats 24 and 1 can't have two boy neighbors
what are the buckets/pigeonholes here?
try to continue filling it in like this
More-or-less the same I suppose?
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Can someone explain if in the last step, the solution, could it be written as -3sinx³ instead of -3x² as the constant is put in the front , like with 10 here.
I do the formula differently(2nd pic) so naturally the values are placed differently compared to the 1st pic.
I'm stuck pls help
,rotate
How'd u get this exact pic to rotate, does it do it automatically
no it can't be written like that
Why not?
the 3 is a constant and can be moved but x^2 is not
it arises from the chain rule while differentiating x^3
so writing -3sinx^3 would be wrong
What about 10?
because 10 is just a number, so it’s a constant
you can move constants around not variable dependent factors
3 is too, I don't understand
yes 3 is a constant but x^2 is not
Yeah but to obtain 10, 5 & 2 were multiplied which have x⁴ & sinx⁵ as coefficients
Which aren't constant
10 comes from constant × constant (2 x 5) right?
Hmm, i see
3x^2 comes from constant × variable so only the constant part can be moved around
It can be moved around? 3? How
“moved” just means rearranged in the product
As in how in this problem?
$-3x^2\sin^2(x^5)\sin(x^3)\quad \text{or}\quad -3x^2\sin(x^3)\sin^2(x^5)$
anflo
or $-,\sin^2(x^5)\sin(x^3),3x^2 ;\text{or}; -3\sin(x^3)\sin^2(x^5),x^2$
anflo
is the fourth one what you mean?
WHAT, i asked if it can be written like that -3sinx³
You said no ..
Yea
i thought you were removing the x^2 my bad
as long as you don't do that its good
So constant can go anywhere regardless
yeah
Hmmm
also is this incomplete?
ah okay
How're you so good ar maths
It's hard to get answers to such minor questions w/o a teacher so this server is yhr last resort sadly
tbh asking these kinds of questions is how it starts
the rest is just practice and time
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I have two answers: 123° and 66°.
For 123°, if we look at the top left corner, the lower angle is 90° - y, and the upper angle is 90° - x. In a rectangle, and then 33° is the middle angle. These angles form a right angle according to rectangles. So, 90° - y + 90° - x + 33° equals 90°. Adding x and y from the left, subtracting 90° from the right, we get 90° + 33° equals x + y, which is 123.
For 66°, we use the alternate interior angles, which means that y and x are both 33°. Adding both of them gives us 66°.
123 is right
okay thanks
alt int angles requires parallel lines
in rectangles each opposite side is parallel no?
oooh wait
i just noticed
it would be 33 + 90-x
for alt int you require red and yellow segments to be parallel
ignore how horribly I've shaded it
i would still get the same answer if i did it correctly 😂
i draw rq
👍
your reasoning for 123 here is correct
horrible to draw on phone
that's correct
happens
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How to disprove isomorphism between these two
I have checked 11 invariants so far, number of vertices, edges, degree sequence, existence of circuit, euler circuit, hamiltonian circuit, and bipartiteness
They satisfied all of them
Better start comparing adjacency matrices and showing that no rearrangement of rows can make them equal
The way things keep going with you and these invariants 😂
did you compare inner/outer degseq as well
well no, lemme try that
does the in/out sequence have to be exactly equal?
obviously yes it does
one graph has 3 vertices with 2 in and 1 out degree, the other has 4 such vertices
well there you have it
but these two graphs have the same seq
redraw the first one by swapping the positions of u1 and u3 on the page
although the first graph has a longest path of length 3 while the second graph has the longest path of length 2, so they are not isomorphic
yes or that
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imagine coloring all the edges in K_10 red or blue
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By examining the chest X-ray, probability that T.B is detected when a person is actually suffering is 0.99. the probability that the doctor diagnoses incorrectly that a person has T.B. on the basis of X-ray is 0.001. in a certain city 1 in 100 persons suffers from T.B. A person is selected at random is diagnosed to have T.B. What is the chance that he actually has T.B.?
can someone draw probability tree for this and show pls
idh my mobile so i cant show my working
no
@serene fog Has your question been resolved?
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what have u tried
i drew a diagram
send
huh
Fijix
can u use different variable for the 2 surfaces
both ur frictions are F(r)
and also how can u say that F(r) = u1 R?
this is only limiting case
fix the notation
ohk then its correct sorry i didnt read limiting eqm
dw dw
i'll redraw the diagram 😭
the diagram's fine
is it the Frs?
yeah its the F_r labels
i dont think u need any more equations..u just have to substitute stuff from here
u can use small f
yeah that works or just F_a or F_b is good too
so my 1/2 mg cos theta = Rb sin theta + Fr Cos theta equation is right?
yeah
so then FrB cos theta turns into u2R cos theta?
mathematically, its right but just a heads up make sure you're visualizing the perpendicular distances for those moments it's super easy to accidentally swap sin and cos if you treat them like force components instead of distances
lol yeah it can be brutal sometimes
once the moments are written with the correct lever arms try eliminating R_a and R_b
if the forces disappear and only θ is left, you’re on the right track
i'm stuck again 😭
oh, what's wrong?
wait maybe i've got it 😭
i'll come back if i get it wrong
Did I make a silly mistake?
@iron estuary
oh you’re really close just a small slip
nooo 😭
in the force balance $R_A \neq R_B$
anflo
when eliminating $mg$ use $R_A = mg - \mu_2 R_B$ not $R_B$
anflo
fixing that gives the missing $\mu_1 \mu_2$ term
anflo
yes yes you got it
where the hell is u1 😭
replace R_a using vertical equilibrium, then simplify
it'll come dww don't stop
what does this mean? have I been using horizontal eq?
horizontal equilibrium is just $R_B = \mu_1 R_A$ right?
anflo
vertical is $R_A = mg - \mu_2 R_B$
anflo
i've made this harder by not including Ra and Rb haven't i?
well i wouldn't say harder... you just took a slightly longer scenic route
man i don't wanna do this anymore, i've got 5 more of these 😭
oh no where are you stuck
I’ve gotten here idk if that Rbs and Ras are correct but idk how to get rid of that Ra at the beginning
sub this
this is just substitution whiplash
you’re allowed to use mg temporarily if it helps eliminate R_a
think of it as changing variables not undoing work
ok then (1/2 (mg) - u2Rb)/u1Rb?
yep that idea just keep mg u2Rb and together in brackets 😭
after that i promise it simplifies cleanly
use your horizontal balance
Am I close? 😭🙏 @iron estuary
$mg = R_A + \mu_2 R_B$
anflo
what have you written?
i wrote mg = u2Ra + Ra...
well its u2Rb right 😭
for this you have to equate torque as well right?
what the hell is torque? 😭
i might be wrong then
this is making me wanna cry 😭
looking at this, mg = Ra + Frb no? and Frb = u2Ra
@iron estuary
what i was thinking was that u have two variables r1 and r2 so to solve them youd need 2 equations
have you seen what i've done so far? apparently it's right and i'm close but i just don't know how to get there
i've thought about vertical equilibrium and horizontal
used Fr = Fmax = uR
and took moments around A
which is against the floor
wait im stupid my bad
$F_{rB} = \mu_2 R_B$
anflo
this is where i've gotten to most recently
if that's true then i've done the whole question wrong 😭
yeah this is correc
you tell me this now 😭
i thought you saw the screenshots 😭
literally says in top right corner 😭
you did everything else correctly looks like to me
OHH
i'm going crazy, i just wanna go to sleep 😭
do i have to do the whole question again from the start?
im so sorry 🙏
nooo it's not your fault at all 😭
well whatever you've done so far is right luckily
i should be able to do this
btw torque is basicaly moment sorry i confused you
oh yh i'm using that
if you sub this you get the answer
u1Ra
but Fra = Rb
because friction is coefficient of friction x normal force at that point
if left = right
horizontal equilibrium gives Rb = Fa = u1Ra but that doesn't mean Fa = u1Rb
should i just do the whole question again but with the right equations?
aa im so sorry i feel like ive caused all this 😭
nonono there's two people helping me now and i still don't fully get it 😭
more practice, it's okayy
if you don't get the idea behind it you should start over
ok 💔
i'll actually read everything this time omg 😭
did you not the first time ⁉️ 😭
i didn't see the right side, well i did kinda but i just didn't notice the A and B
okok 😭
its okay it'll be quick quick
What is the issue right now for other helpers?
some channels have like ten of them rotating in and out
but usually courtesy is that if a channel already has good progress with a helper we don't step in unless to correct mistakes
really ten?
well there was a big mistake but it's ok 😭
😭
anflo's doing a great job
yes for long channels or channels where everyone feels the need to step in for whatever reason
really i appreciate that but it was my fault
shhh, it's ok
this week there was one with 20
the coordinate system guy
oh wow
ohh Phoenix?
nah HereC
ok let's get the channel back on track
well, i hope i'm not the next
Well this channel has had 300+ messages so far and it seems like the problem is still unresolved
-# mathematics should delete itself he said 
oh dear, we're getting there 😭
just remember friction is μ × normal at the same contact point
stop lying 😭
now you can sleep peacefully
i'm contemplating if i want to so another question 😭
yh lol
next one will feel way shorter after this
nah genuinely
the next question had three parts 😭
6, 3 then 8 marks
okayy, can i find you if i need help again? 😭
sure
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Given a pyramid S.ABCD with base ABCD being a trapezoid with bases AD and BC, let M be the midpoint of SA. Knowing that BM is parallel to the plane (SCD), calculate the ratio AD/BC.
I know I have to use Thales theorem but I just can’t prove it
@turbid salmon Has your question been resolved?
Have you made a diagram?
If so, send it here, else make one and send it here
try making the base lie in the plane z = 0, and make abcd coordinates
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in this problem's solution: https://wiki.randommath.com/aime/2017/part-i/problem-15
Review the full statement and step-by-step solution for 2017 AIME I Problem 15. Great practice for AMC 10, AMC 12, AIME, and other math contests
what are you asking
oh give me a sec
i don't understand how they derive the A
like capital A
i know what it's helpful, but idk what they did to get A
(Denominator of s)
yea but like the coefficnets of cos and sin are different
Draw a triangle
Hypotenuse A
One side 7sqrt(3)
The other 11
By pythagorean thm, A = what they write
well yea i know that
Mark the angle adjacent to the 7sqrt(3) side as alpha
Then cos(alpha) = 7sqrt(3)/A and sin(alpha) = 11/A
wait i'm dumb
This gives us the rewriting of the fraction
.close
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hello i need help with factoring, i dont understand how im supposed to know which numbers are the right ones to use if that makes sense.
for example i wanna use 7,8 to factor here but ik it will be smaller numbers but im so confused how im supposed to figure them out
hopefully that makes sense
you need to find the largest square that divides 56 
how am i supposed to know what the smallest square is..
is that something you just memorize
(well, technically the largest square that divides 56)
whoops, I dunno why I said smallest 
trial and error is pretty effective when you have small numbers
I am guessing you wanna know how to divide 56 into its prime factors? like how to write 56 as 2x2x2x7?
for instance, it can't be 7 because 7^2 = 49 doesn't divide 56
so it's gotta be smaller
the only thing is i try to do trial and error but its hard when i dont know whats right or wrong
so you want to know whats the largest number that you could take outside the sqrt?
ie the largest square factor?
wait why 2?
in that case write the prime factorization of the number that you have
YES THIS
oh, prime factorization is a bit different 
so you said that 56=7*8
right
now 7 is prime, but 8 isnt
so write 8 as a product of primes too, how does that look like?
so that means i need to go lower?
okay give me a sec i rlly dont know what primes are either😭
yes, you need to have a product of only prime numbers
I'd recommend having a look e.g. at https://youtu.be/XBnUWjo3TgM
Welcome to Prime Factorization with Mr. J! Need help with how to find the prime factorization of a number? You're in the right place!
Whether you're just starting out, or need a quick refresher, this is the video for you if you're looking for help with prime factorization. Mr. J will go through prime factorization examples and explain the steps...
np, a prime number is a number whose divisors are 1 and itself
so that will tell me what primes numbers are?
-# And also learning what prime numbers are, since that's paramount to what the goal here is
ohh so basically like the lowest?
so like a natural number p is said to be prime if the only divisors of p are 1 and p
lets have some examples
so basically like 2 and 3 because they cant get lower without 1?
for example the number 13, the only numbers that divide 13 are 1 and 13 right?
so 13 is prime
-# Okay, to paraphrase Inigo Montoya, you keep using that word [lowest] but I don't think you're quite clear on what you mean by that
okok i see
on the other hand take 12 for example, then its divisors are 1,2,3,4,6 and 12
so it isnt prime
so basically what i would need to do, if i assumed 7,8 could work, i need to test to see of both are prime numbers, if not i go lower?
-# [you've missed 4, but yes]
i noticed that and knew that you will probably correct me (since i saw you typing right after i sent my message) so i tried to be fast but you were faster
wait so since i wanted to try 7,8 since 7 is a prime number but ik that 8 isnt, wouldn’t i just be able to break down 8
but keep 7?
yep
yea, now what does 8 break up to?
right
OKOK i get it but
so 2 is the only prime divisor
and it occurs 3 times
so 2^3. Now to answer your question, find the largest square in the factorization 7*2^3 of 56
so like here, how 5 stayed the same, 7 would also stay the same correct?
okok
exactly
is 7 the only thing that stays the same tho?
no the z^7 aswell right?
you said that 56=7*2^3 and that 7 is a prime so it certainly isnt a square
to find the largest square that’s something ill just have to memorize aswell for future questions?
right
so prime numbers cant be square?
now what about 2^3? is 2^3 a square? if no then whats the largest square in 2^3?
right, because if p is a prime number and its a square then it can be written as p=a^2 for some a y definition of square right?
8 since it = 8 right?
but then a can neither be 1 nor p because 1^2=1 and p^2 is different than p, which would mean that a is a third divisor of p and thats impossible
at most you will have to memorize list of prime numbers which are easy to recognise for small numbers problems only arise when you reach big numbers which look prime but arent like 57
i am not sure what this means
/genuine - Do you know what a square number is?
what is the meaning of a square number
2^3 prime is 8 cus it 2^3 =8
not really 😭😭
ive looked it up on a couple past questions so ik like a few
have you seen the term perfect square before?
i think so, when i was breaking down past example questions
a square number is the same as a perfect square, i just used a different name. Anyways, a square number/perfect square is a number that can be written in the form of a^2 for some a
like 25=5^2, 9=3^2, 49=7^2 etc...
ohhh i see, but to confirm “a” in this situation is the perfect square right?
so 5 3 7 are the perfect squares?
or id it 25 9 49 that are the perfect squares?
It'd be dangerous to try and define "square number" using "squaring"
A square number is a number that is the area of a square with unit-lengths
So for example, if you have a square of side-length 3, it has area 9
So 9 is a square number
but doesn’t that mean anything could be a square number?
Why do you think that?
because you say it’s anything that is side length, couldn’t any number be used for a square?
Any whole-number side-length, yes
But the square number is the AREA there
Not the length
Notice that I never said THREE was a square number here?
but if i did area of square as 4 wouldn’t that mean 16 is also a square number?
hmmm you must have a downside for this in mind to say this but i am not sure what makes it harder to define it like this
(because you ought to define what "squaring" even does)
i see, fair point
4 is a square number - it's the area of a square with sidelength 2;
similarly 16 is a square number since it's the area of a square with sidelength 4
So yes, both 4 and 16 are square numbers
wait
a sqaure number is a x a where a is a natural/counting number so yes a square of a square number would also be a square, like for a=4 a*a=16 so 16 is a aquare number
so does that mean 25 is also a square number?
i always avoid interpretting things geometrically thats why i wouldve never thought about defining it in this way
yes
SO ANYTHING TIMES ITSELF IS A SQUARE NUMBER?
yes
ye
(it at least explains why we call this "squaring" in the first place, though you have a point there)
a square number is basically that a number which you can get by multiplying a number by itself
omgg that makes everything so much easier
yea thats definitely true and is a good way to explain it for the first time 
so square number and perfect square are the same thing correct? not 2 different terms?
yep
so there is no square in 2^3?
If you write it as 2 x 2 x 2, ...
Can you spot where we're multiplying the same number with itself?
2 x 2 IS the square number
that just confuses me now
right but we cant just forget abt the other 2
what am i supposed to do with it
right
When we write a x a to mean a times a, THIS we can rewrite as a^2
and 2^2 is the square number
oh I see you meant the whole 2x2x2 bit
yh that works
So you've got 7 x 2 x 2^2 - there any more squares?
but now that i know this how does it factor unto the equation? or did you ask me this to check that 2^3 was correct?
wait why cant i write it as 2^3 x 7
-# [@uneven knoll it's in part because this was the goal...]
thats what you have reached from all of the work done above
yep
you should look at 2^3 first
wait so 2^3 goes lower?
2^3=2 x 2 x 2
you mean thats why you chose to define a square number in terms of the area of a square?
doesnt it just end as 2?
can you break it down in a way that leads to a sqaure number?
so at the end i just remove the ^3 and left with 2 correct?
usually i do that at the end
nit exactly
i am guessing you are saying root(2^3)=2?
honestly no, i just noticed from the videos and examples they just would remove the ^2 at the end
i dont actually know the reason behind it
right
you can see that there is a square number hiding in 8 what would it?
its 4?
yes!
so can you write root(8) as root(4) x root(2)
so i can write it as 4^2?
no thats 4x4
can you explain more what you mean by 2 separate numbers?
like this?
yes so now what is root(4) ?
2 right?
2?
thats what you get outside the root
ohh 4
okay now im lost
see root(56)=root(4) x root(7) x root(2) you got till here right?
right
now you know root(4) is just 2 so you write is at 2
but you cant forget about the non square numbers
wait now im confused
you may not be able to write them outside the root but they still exist
so after this i just do 7 x 2?
so root(4)=2
root(56)=2xroot(7)xroot(2)
yes but you keep them under the root
im confused what keep them under the root means
its like you keep root(2) as root(2) as it isnt a perfect
you cant write it as 2 without the root
im still confused
what do i do with the equation then
so i should or shouldnt write it as 7x 4x 2
you can write root(56)=root(7x4x2)
but how do i solve it
but 4 is a perfect sqaure so you took it out as a normal number that is 2
so i got the root part right?
now you are left with root(56)=2xroot(2x7)
what is rot
mb spelling mistake
wait im confused how is 2xroot(2x7) written like
If i might add a detail, you know exponents right?
$16 = 2^4 = 2\cdot2\cdot2\cdot2$
just like here you wrote 2 x root(5)
im just even more confused where is 16 coming from😭
i dont know what exponent rules mean but looking at a picture i know what that is
i know that 16 = 2x2x2x2 = 2^4
if thatswhat ur asking
like:
$$2^3 \cdot 2^2 = 2^{3+2} = 2^5$$
$$\left(2^3\right)^2 = 2^{3\cdot2} = 2^6$$
The second one is important
well kinda, i just relearned it today so its familiar but idk anything beyond the basics
well, square roots are a sort of exponent too
to be specific, we say that $\sqrt 2 = 2^{\frac 12}$
thats why elements that are a perfect square become an integer
im getting more n more confused 😭
$$4 = 2^2$$
$$\sqrt 4 = \sqrt{2^2} = \left(2^2\right)^{\frac 12}$$
and 2 * 1/2 = 1
so you end up with 2^1 = 2
And this not only applies to 2^2, also for 2^4, 2^6, etc...
im really trying to understand but im just getting more n more lost😭
could you rewrite 81 in terms of its prime factorization?
?!?!??
81 is a perfect square btw.
whats that gotta do with my question man im getting confused cus im straying more n more away from what im trying to figure out😭
understanding how square roots work as exponents basically simplifies all of this to a purely symbolical thing
Nope, thats perfect squares.
the square root is an operation that undoes the "squaring" of a number
so then how does square roors factor into my equation
so then square root of 81 is 9
yea
-# you're doing amazing char!!!!!!!
now, prime factorization is useful, because it allows you to easily find the square root
If its an integer number
similar example to your problem:
the square root of 20.
do you agree that 20 is equal to 2 * 2 * 5?
yea
then we can rewrite it as 2^2 * 5
i know
i just learned this
this is what i was confused on
but i got an understanding of it now im just trying to incorporate it into my problem
wont work like that
can you find the factorization of 56? we will then deal with the z^7
Well, then you have to consult this almost by force.
are you confused about what square roots have to do with your problem?
!occupied
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
im confused where square root is being incorporated in the problem, we figured out factorization of 56 i thought itd 7x4x2
or 7x2^3
your entire question is about sqaure roots
i thought it was about square numbers
genuine question you do know that this symbol is the symbol for sqaure root?
i needed help factoring the 56 right, which is square numbers
yes
for easy visualization: $\sqrt{2^2 \cdot 2 \cdot 7} = \sqrt{2^2}\cdot\sqrt{2\cdot7}$
I hope it eventually makes sense why i divided it like that
<@&268886789983436800> troll in help channel
so why can i not just use 2^3 x 7 if they equal the same thing?
I've already redirected you to #❓how-to-get-help if you're actually genuine about wanting help. eitherway, the mods will probably warn you 
Remember the fact that the square root undoes squaring?
Thanks
means, if you got $\sqrt{2^2}$, thats just equal to $2$
right
they were... banned?

