@humble hollow Khan Academy and brilliant.org are great resources. Brilliant requires a paid subscription, but it will definitely push you like you want it to.
#help-27
rain dock
lavish fog
hi
if sqrt(x^2) = |x|
why is sqrt((-1)^2) = -1
and not 1
devout snowBOT
hi
if sqrt(x^2) = |x|
why is sqrt((-1)^2) = -1
and not 1
restive river
is sqrt((-1)^2) equal to -1?
restive river
why?
lavish fog
is sqrt((-1)^2) equal to -1?
idk man
I remember vids saying it is
but I think it should be positive 1
so ya im kinda confused
am I wrong? or is the vid wrong
restive river
I think your videos or how you interpreted them/remember them may have had an issue
sqrt((-1)^2) is 1
main gull
You might be mixing up
$\sqrt{(-1)^2}$ and $\sqrt{-1}^2$
Because $\sqrt{(-1)^2} = 1$ while $\sqrt{-1}^2 = -1$
woven radishBOT
dldh06
lavish fog
You might be mixing up
$\sqrt{(-1)^2}$ and $\sqrt{-1}^2$
Because $\sqrt{(-1)^2} ...
ah
I am
but arent they the same?
main gull
Nope
lavish fog
can u explain why?
cuz as far as I know
((x^a)^b) is the same as ((x^b)^a)
main gull
Order of operations
Type it into a calculator and see for yourself
,calc sqrt((-1)^2)
woven radishBOT
Result:
1
,calc sqrt((-1))^2
woven radishBOT
Result:
-1
be careful with those, exponent rules don't apply for every number
lavish fog
Order of operations
hmmmm
is it diff for negative numbers?
because
((x^a)^b) is the same as x^(ab)
and ((x^b)^a) is the same as (x^(ba))
and yknow
multiplication is communtative
lavish fog
be careful with those, exponent rules don't apply for every number
hmmmm
ah yea
that square root multiplication thingie for negatives
are there anything else?
restive river
exponent rules don't apply very nicely for exponents on negative numbers
they aren't even rules in that case
cinder topaz
devout snowBOT
cinder topaz
im confused on the set up
for #2
i tried to put the variable for edwin and the. for rhea but it just doesnt make sense to me
<@&286206848099549185>
devout snowBOT
@cinder topaz Has your question been resolved?
devout snowBOT
@cinder topaz Has your question been resolved?
weary narwhal
There's no way I'm doing this right... Because the answer is fairly far off when I try and calculate it afterwards. The only number I could think of that even comes close to the bases are 2 and 4, and the result I get is quite off. Does anyone have some advise to point me in the right direction?
devout snowBOT
There's no way I'm doing this right... Because the answer is fairly far off when...
main gull
There's no way I'm doing this right... Because the answer is fairly far off when...
How off are you?
weary narwhal
I'm accurate to 1 decimal place
so, I get something around 58, not 60, when I raise 15 to the power
main gull
What if you started at 8 instead? Doing log base 8, then making that into log base 4 then log base 2
It's more work but could be more accurate
opaque bane
devout snowBOT
opaque bane
.close
hollow comet
Help
hollow comet
stuck field
Which one?
hollow comet
Which one?
How to do the bottom right?
stuck field
Multiple ways.
Take components.
Or just add the two vectors.
What's the angle between the two vectors?
hollow comet
So it 4 and 122 degree?
stuck field
4 what?
stuck field
Lol
hollow comet
How do you do it?
stuck field
Take components.
Or directly add.
As the angle is 122°
You just need the magnitude right?
hollow comet
Idk it just algebra
stuck field
Ye just algebra.
hollow comet
Ima just do it and sho you the answer
stuck field
That works.
hollow comet
-2.11R+9.39U
hollow comet
That works.
The answer
stuck field
R is right U is up?
hollow comet
R is right U is up?
Yes
stuck field
Seems incorrect to me.
hollow comet
What’s your answer?
hollow comet
Let me do it.
Ok mate
stuck field
Alr
My answer's diff.
Do you know the answer?
Correct ans?
How'd you get that answer
stuck field
Seriously though. How'd you get that ans anyway?
stuck field
3.39R* is infact correct.
hollow comet
Sin is 6U and cos is 0R
stuck field
Correct.
hollow comet
So is it ?
hollow comet
And use that for 32 too.
But it 122?
stuck field
Sin is 6U and cos is 0R
This seems correct.
stuck field
But it 122?
Ah. It is, then again you're solving for them separately. 32 is the angle between axis.
I see that's why you got that answer.
Let's just name horizontal axis as x.
And vertical y.
That'll be easier.
Then figure x and y components for each one separately add them ultimately.
hollow comet
8.11U, 3.39R
stuck field
By taking components, I mean the same thing you did for the first one.
R is correct.
U is incorrect.
hollow comet
What is U then?
stuck field
U must be less than 6.
3.88U + 3.39R
Let me elaborate,
You took 90° for the 6 unit force.
hollow comet
3.89u and 3.39
stuck field
3.89u and 3.39
Yea
stuck field
You're welcome.
devout snowBOT
stuck field
That I blame you xD
I am leaving the country anyways mate.
devout snowBOT
Available help channel!
distant minnow
devout snowBOT
distant minnow
should i prove this using pumping lemma ?
if yes how to do so ?
and is there any another way apart from pumping lemma to check whether a language is regular or not ?
<@&286206848099549185> !
rain dock
I'm not familiar with the terminology. What does it mean for a language to be regular
I think I may be of use. But if not, I recommend you ask in #proofs-and-logic . It'd be easier to find help on the topic there
devout snowBOT
@distant minnow Has your question been resolved?
devout snowBOT
Can anyone help.my friend with math
green fjord
We are in 384kbps
devout snowBOT
@green fjord Has your question been resolved?
devout snowBOT
@green fjord Has your question been resolved?
faint zinc
@green fjord #❓how-to-get-help ask your question if you want it answered
green fjord
Nm
last umbra
devout snowBOT
last umbra
what test would you apply?
supple knot
anytime you have (-1)^n, alternating series test should be one of the first ones you try
last umbra
riemann always here in my time of need
devout snowBOT
Available help channel!
restive river
How can I find the shortest possible length of the hypotenuse when x is is greater than or equal to 0 or less than or equal to 1?
devout snowBOT
How can I find the shortest possible length of the hypotenuse when x is is great...
@restive river Has your question been resolved?
lime linden
What's the probability of obtaning 2 tails in 5 coin flips?
wicked turtle
so i was right?
what was your answer? you wrote two things, 1/6 and 5C2
the latter is obviously not a probability
you also need to clarify the question: do you mean exactly 2 tails or at least 2 tails?
lime linden
what was your answer? you wrote two things, 1/6 and 5C2
thats my work
the latter is my work
lime linden
you also need to clarify the question: do you mean *exactly* 2 tails or *at leas...
thats just what the q said word for word
lime linden
if its at least then it would be 5c0 + 5c1 + 5c2/ 2^5 right?
lime linden
1/6 is not correct
then what is it then?
wicked turtle
if it's exactly 2 heads then (5C2)/32
lime linden
you mean tails?
wicked turtle
sorry yes tails
lime linden
but shouldnt make a difference in the calculation
wicked turtle
2 heads, same answer
lime linden
oh i forgot the divide by 32
lime linden
if its at least then it would be 5c0 + 5c1 + 5c2/ 2^5 right?
but this is correct tho right?
lime linden
if it's exactly 2 heads then (5C2)/32
1/192
wicked turtle
5C2 is an integer
it's not 1/6
more generally, N choose K is always an integer (assuming 0 <= K <= N)
lime linden
it's not 1/6
what is it than
wicked turtle
5 choose 2 is (5! / (2! x 3!)) which is (5 x 4) / (2 x 1) = 20/2 = 10
lime linden
i mightve been doing binomial coefficient wrong this whole time
wicked turtle
oops 😁
wicked turtle
yep
lime linden
ik now
wicked turtle
yep
lime linden
if its exactly 2 tails
wicked turtle
correct
lime linden
if its at least then it would be 5c0 + 5c1 + 5c2/ 2^5 right?
this would be at least 2 tails right?
this would be right for that?
wicked turtle
assuming you meant (5c0 + 5c1 + 5c2) / 2^5
all three are over the same denominator
wait
that's at most 2 tails, not at least
at least 2 tails = 2 tails or 3 tails or 4 tails or 5 tails
so (5c2 + 5c3 + 5c4 + 5c5) / 2^5
lime linden
that's at *most* 2 tails, not at least
oh yeah
at least would be from 2 to 5
right?
5c2 to 5c3 to 5c4 etc
lime linden
at least 2 tails = 2 tails or 3 tails or 4 tails or 5 tails
yup
wicked turtle
you can also do it as (1 - at most one tail), which is 1 - (5c0 + 5c1)/2^5
which is less to calculate
lime linden
you can also do it as (1 - at most one tail), which is 1 - (5c0 + 5c1)/2^5
for least?
wicked turtle
yes
deep flower
```What's the probability of obtaning 2 tails in 5 coin flips?```
atleast or exactly 2 tails?
deep flower
$P(A)= \frac{n(A)}{n(S)}$
woven radishBOT
2022 squared
lime linden
what?
deep flower
why 5C2?
wicked turtle
n(A) = 5C2, n(S) = 2^5 in this problem
lime linden
n(A) = 5C2, n(S) = 2^5 in this problem
yeah
What's the probability of getting at least one six in two rolls?
this would be 7/36 right?
since theres six ways to get 1 six, and 1 way to get 2 six right?
this is more of a counting q right?
devout snowBOT
@lime linden Has your question been resolved?
lime linden
<@&286206848099549185>
lime linden
anyone?
wicked turtle
```What's the probability of getting at least one six in two rolls?```
two dice rolls?
that's equal to (1 - probability of no six)
lime linden
yes
wicked turtle
and i bet you can find probability of no six pretty easily
lime linden
wait how?
fuck
i misread the q
u got 11/36 right?
my original answer would be correct if its at most 2 six right?
im gonna stop doing math
my brain is messed up
wicked turtle
probability of at most 2 sixes in 2 rolls is 1
(you can't have more than 2 sixes, right?)
probability of no six is (5/6)^2, which is the probability of no six on the first roll times the probability of no six on the second roll
which equals 25/36
therfore probability of at least one six is 1 - 25/36 = 11/36 as you said
devout snowBOT
@lime linden Has your question been resolved?
lime linden
my original answer would be correct if its at most 2 six right?
@wicked turtle ?
devout snowBOT
@lime linden Has your question been resolved?
worthy harbor
p(at least 1) = 1-p(none)
worthy harbor
therfore probability of at least one six is 1 - 25/36 = 11/36 as you said
I got this answer @lime linden
lime linden
I got this answer <@910344998397050950>
same
but im asking if 7/36 would be if its at most 2 six right?
worthy harbor
at most 2 sixes is 2/2 sixes, which is accounted for in 1-p(none)
in other words, we can say p(at least 1 {out of 2, conditionally based on the problem}) = p(1)+p(2)
p(1) is calculated to be 10/36, and p(2) is calculated to be 1/36, adding these two together gives us 11/36
makes sense @lime linden ?
lime linden
ik bruh
im asking if 7/36 would work in a situation where its AT MOST 2 sixes in two rolls
worthy harbor
I understand taht
now tell me this: what are you asking for with at most 2 sixes? what probabilities are you adding up
lime linden
so if we roll both die twice we can 1 6 or both of them to be 6
so in that case
for 1 6
theres 6 options for that
(1,6) to (5, 6)
than we have (6, 6)
worthy harbor
okay you are right in a case
however, (1,6), (2,6), (3,6), (4,6), (5,6) is 5
and then you have (6,1), (6,2), etc.
lime linden
yeah so 6/36 not 7/36
lime linden
does order matter?
lime linden
so if order matter than (6, 1) doesnt count than
worthy harbor
no it does count
lime linden
oh
worthy harbor
if you were counting the probability of getting at least one 6 in two rolls, and you got a 6 and a 1, you wouldn't throw out that option
after all, you got a 6 in that set
lime linden
11/36 for at most than right?
lime linden
oh so at least and at most are the same
worthy harbor
because p(at least 1) out of 2 rolls = p(1) + p(2) AND p(at most 2) = p(1) and p(2)
worthy harbor
oh so at least and at most are the same
not in every case
worthy harbor
not only in this one either
lime linden
but in other ones
worthy harbor
yes there are other cases
lime linden
alr thanks
worthy harbor
i have a question tho
have you learned binomial formula?
,,P(X=x)={n \choose x}(p)^x(1-p)^{n-x}
worthy harbor
woah
woven radishBOT
JWCfive
worthy harbor
Because the binomial formula is EXTREMELY useful in these problems
what I did to solve the problem was plug in 2 for n (n is the total number of tries), 1/6 for p (p is the probability) and then 0 for x (x is the number of sucesses)
worthy harbor
nCr is used in the formula
lime linden
within?
worthy harbor
yes
worthy harbor
,,{n \choose x} = nCx
woven radishBOT
JWCfive
worthy harbor
Be sure to check the conditions for binomial formula to be used:
-
Binary - there are two outcomes, success or failure
•in this problem, success = one 6 or two 6 -
Independent - the outcomes do not affect the probability of the other outcomes
•in this problem, rolling one 6 does not make rolling another 6 less likely -
Number of successes - there is a number of sucesses, rather than a number of times until sucess
•in this problem, we find the probability of 0 sucesses -
Same probability - every event has the same probability
•in this problem, the chance of rolling a 6 is always 1/6 (similar to independent)
Also, another way to think of the formula:
,,P(number of successes) = {total number of trials \choose number of successes} * (probability of success)^{number of successes} * (probability of failure)^{number of failures}
lime linden
thats a simplified version of binom coefficient right?
worthy harbor
kind of
its pretty much the same actually
well binom coefficient is used in the binomial formula
lime linden
ye
woven radishBOT
JWCfive
worthy harbor
This is for the conditions...
-
Binary - there are two outcomes, success or failure
-
Independent - the outcomes do not affect the probability of the other outcomes
-
Number of trials - there is a number of trials until success
-
Same probability - every event has the same probability
wicked turtle
@lime linden sorry, i was afk and saw you asked about rolling "at most two sixes in two rolls" - I replied earlier but you probably missed it:
probability of at most 2 sixes in 2 rolls is 1
(you can't have more than 2 sixes, right?)
lime linden
yeah
worthy harbor
do .close to close
devout snowBOT
@lime linden Has your question been resolved?
devout snowBOT
devout snowBOT
Available help channel!
trail pulsar
hello
zenith jacinth
with what
trail pulsar
trigonmetric function
maximun point and mid
find the midline of it
@zenith jacinth
this shit is hard
amber scroll
What
trail pulsar
can you solve it
narrow plank
What's the maximum value of the function (highest point)?
What is the y-value that is halfway between the highest and lowest point? @trail pulsar
trail pulsar
i don't get it
the max value is (1/4n,-2.2) and min (3/4n,-8/5)
@narrow plank
why is math so fucking hard
that is the main question
why is math math
who created math
narrow plank
What is the y-value that is halfway between the highest and lowest point?
narrow plank
What don't you understand from that statement?
earnest stirrup
who created math
we did
devout snowBOT
@trail pulsar Has your question been resolved?
lilac heath
lilac heath
you know the equation for a line?
earnest stirrup
@trail pulsar is it a test?
lilac heath
we dont just give out answers
earnest stirrup
you know the equation for a line?
(and right now, Samianz, you are the one whose supposed to give answers btw 🙂 )
lilac heath
np
earnest stirrup
then answer him xd
trail pulsar
bro how do I say this
im the dumbest person
in my fucking class
i don't know math
simple
lilac heath
okok ill draw a diagram for u
trail pulsar
what do I have to find tho
lilac heath
write the equation for it and ur good
trail pulsar
1.05?
lilac heath
-5.35 was right for where the line crosses over the y axis
but its not an equation
theres no equal sign
do you know the equation y = mx+b ?
trail pulsar
yes
lilac heath
use that
lilac heath
y=mx+b is the equation of a line where m is slope and b is the y intercept
trail pulsar
can you dm me the answers
(not breaking the rules)
how about you give me the 1st questino answer
and u can give me the
other one
I*
wait
2?
@lilac heath
lilac heath
y=mx+b is the equation of a line where m is slope and b is the y intercept
like this
lilac heath
do u know what slope is
trail pulsar
no
lilac heath
u will just have to learn then
lilac heath
yeah dont ping that guy so much
devout snowBOT
@trail pulsar Has your question been resolved?
steel comet
devout snowBOT
steel comet
Why have we added a constant here?
steel comet
Do you agree that d/dx(x³/3 + C) = x²?
No
steel comet
It will have a quadratic derivative
Yeah
graceful cosmos
Integration is difficult. You'll want to make sure you are up on your derivatives
steel comet
Integration is difficult. You'll want to make sure you are up on your derivative...
Fr.
So like we add C in order to make sense of the problem?
graceful cosmos
Do you agree that d/dx(x³/3 + C) = x²?
steel comet
Like lesstake x=2... 2^3/3 = 16/3
Then we add a C to it
Wait no that doesent make sense
shy edge
Like explination?
Let's do different
steel comet
Okk
shy edge
What's d/dx(x² + 2)
shy edge
2x + 0 which is simply 2x
steel comet
2x + 0 which is simply 2x
Yep
shy edge
If you don't get the definition of differentiation yet, you should start by that before integrals tho
Now d/dx(x² + 3)
steel comet
If you don't get the definition of differentiation yet, you should start by that...
2x
steel comet
Now d/dx(x² + 3)
2x*
shy edge
Alright
So
Remembering the definition in your book
About integrals
Yes! That's the definition of your integrals
Alright so
shy edge
Now let's say
f(x) = x² + 39722892
d/dx (x² + 39722892) = 2x
And use the antiderivative process to find your way back to f(x)
shy edge
Integral is the antidericative process you gonna use to find your way back
to f(x)
In this case we know that f(x) is x² +39722892
steel comet
In this case we know that f(x) is x² +39722892
Yep
shy edge
Shit is we can't use integral to find the way back if you don't know f(x)
Unless you got further information
Which we do in this case
So let's suppose we got some function f(x)
Which d/dx[f(x)] is 2x
We know that the first derivative of this function is 2x
So we go and take the antiderivative process (also known as integral) to get back to f(x)
And there we go
$\int 2x dx = x²$
woven radishBOT
Akz
shy edge
And well thats wrong
Because f(x) could be just x²
Or x² + 1
Or x² + 837282287
Yet it would be still 2x when we differentiate
So if we don't know f(x) or any further information that could help us
The best we can do is say that
$\int 2x dx = x² + C$
woven radishBOT
Akz
shy edge
Cause when u gonna differentiate x² + C, c can be any number and yet it's still will vanish and become just 2x
crimson vault
f(x) is a continuous function
devout snowBOT
f(x) is a continuous function
crimson vault
$\int _{\frac{1}{n}}^n:\left(1-\frac{1}{x^2}\right)f\left(x^2+\frac{1}{x^2}\right)dx$
woven radishBOT
Serky
devout snowBOT
@crimson vault Has your question been resolved?
devout snowBOT
low holly
If you let $x =$ that sum, recognize that $x = \sqrt{20 + x}$
woven radishBOT
tushar
low holly
From here you can solve the quadratic and extract the only positive solution (since the RHS is clearly positive)
restive river
how to solve area for this ?
I thought it would be 4 + x times 3 + x then minus x squared for the quarter that's gone but it doesnt seem to be the answer
feral scarab
You can seperate the shape and find the area of the two
rigid lodge
devout snowBOT
rigid lodge
Inconsistent results for some reason
Excel gave the same result
Wait
Egg on my face
Used the wrong function
.close
bold prism
How do I show that $$\newcommand{\R}{\mathbb{R}}
f \colon \R_{+} \to \R_{+}, f(x) = x^2$$
$$
\mathbb{R}_{+} = [0, ∞)$$ is bijective
devout snowBOT
How do I show that $$\newcommand{\R}{\mathbb{R}}
f \colon \R_{+} \to \R_{+}, f(x...
young plover
what have you tried?
bold prism
I proved that it is injective but I don't know how to prove it is surjective
young plover
well you need to show that for all $y\in\mathbb{R}+$ there exists an $x\in\mathbb{R}+$ such that $f(x)=y$, right?
woven radishBOT
llspacebarll
young plover
so what would x have to be for any given y?
bold prism
that would be y^1/2
bold prism
so thats it?
woven radishBOT
llspacebarll
bold prism
I remember my teacher telling me to prove that range of the function should be equal to the codomain :hmm:
bold prism
so range of y^1/2 is R_+ too right
young plover
then you'd have to show that $f(\mathbb{R}+)\subseteq \mathbb{R}+$ and $\mathbb{R}+\subseteq f(\mathbb{R}+)$
woven radishBOT
llspacebarll
bold prism
**llspacebarll**
f(R+)?
young plover
the image of $\mathbb{R}_+$ under $f$
woven radishBOT
llspacebarll
bold prism
is that how your write range?
young plover
yeah
bold prism
oh alright
bold prism
**llspacebarll**
ok then I guess this is easier
young plover
yeah probably
bold prism
ok I had another question, how would I prove f(x) = x^3 + x^2 is not injective but surjective
young plover
what's the domain?
bold prism
R -> R
young plover
well then there's your hint in and of itself
its not just the positive reals now its all of R
so use that to your advantage
you just have to show that for one pair $x,y$ where $f(x)=f(y)$ you have $x\neq y$
woven radishBOT
llspacebarll
young plover
you are free to choose x and y as you please
bold prism
counter example right
young plover
yep precisely
young plover
definitely dont guess randomly
you need to think how can you get two different numbers to equal the same thing after applying the function to them
i think negatives will be your friend
since the square will remove the negative
bold prism
ah x1=-1, x2=0
bold prism
f(x1) = f(x2) but x1 != x2
young plover
exactly
bold prism
that's done how about onto
y=x^3 + x^2
i'll need to express x purely in terms of y right
x= ...
how do I do that
young plover
this one isnt immediate for me, gimme a bit to think
bold prism
alright
bold prism
y=x^3 + x^2
i'll need to express x purely in terms of y right
x= ...
is there even a way to do this?
young plover
no, i think this one might be easier if you show equality of range and codomain
bold prism
alright how do I find the range of y=x^3 + x^2
young plover
i would do it like this:
sorry lemme type here for a sec
$y=\frac{1}{2}y+\frac{1}{2}y=\frac{1}{2^{2/2}}y^{2/2}+\frac{1}{2^{3/3}}y^{3/3}=\left(\frac{1}{2^{1/2}}y^{1/2}\right)^2+\left(\frac{1}{2^{1/3}}y^{1/3}\right)^3$
woven radishBOT
llspacebarll
young plover
and so for any real number $y\in\mathbb{R}$ you can write in the form of the function and thus $f(\mathbb{R})\subseteq \mathbb{R}$
woven radishBOT
llspacebarll
bold prism
**llspacebarll**
young plover
lol
bold prism
How did you do that
young plover
do you need clairfication?
bold prism
yes
young plover
well the first step you can agree with right?
bold prism
do surely didn't do it from left to right
young plover
i did
bold prism
so how do we find the range
of x^3 + x^2
should I close this channel and create a new one as the question is diff
I think I should
.close
high nebula
For any $a, b \in \mathbb{Z}$ with the condition that $a \neq b$ and $\sqrt{a}, \sqrt{b} \not\in \mathbb{Z}$, prove, or give a counterexample, that $\sqrt{a} + \sqrt{b} \in \mathbb{R}- \mathbb{Q}$ and $\sqrt{a} - \sqrt{b} \in \mathbb{R}- \mathbb{Q}$
devout snowBOT
For any $a, b \in \mathbb{Z}$ with the condition that $a \neq b$ and $\sqrt{a}, ...
woven radishBOT
Taiga
devout snowBOT
@high nebula Has your question been resolved?
devout snowBOT
@high nebula Has your question been resolved?
restive river
For any $a, b \in \mathbb{Z}$ with the condition that $a \neq b$ and $\sqrt{a}, ...
hello - do you have any ideas yet?
high nebula
i've tried to use proof by contradiction, but didn't get very far. i only know that both can't be rational at the same time, but i don't know if you can have a case where one is rational and the other is irrational
restive river
are you worried about one of $\sqrt{a}$ and $\sqrt{b}$ being rational and the other irrational?
woven radishBOT
yume ♡
restive river
if so, it turns out that for any non-negative integer a, sqrt(a) is either an integer or is irrational
high nebula
i'm worried about $\sqrt{a} + \sqrt{b}$ being rational but $\sqrt{a} - \sqrt{b}$ being irrational
woven radishBOT
Taiga
dark tundra
Given that the required thing has the connector “and”
restive river
hmm it says 'and'
dark tundra
If you can give a counterexample that has one rational and another irrational, you’ve disproven the claim
high nebula
yeah
restive river
do you think it is true or false?
high nebula
but we don't know whether it's true or not
it feels like it's true
basically it says that a and b are integers, not equal and not perfect squares
technically they're natural numbers in this case
since if either a or b is equal to 0, then it's obviously irrational for both
restive river
maybe it's easier to go through an example first before trying the general proof. Would you be able to prove that, say, sqrt(2) + sqrt(3) is irrational?
as well as sqrt(2) - sqrt(3)
high nebula
hmm, i could try
restive river
hint: ||I believe if you use euclid's lemma, the proof generalizes nicely||
high nebula
the main question or the one you asked now?
the lemma seems a bit obvious to be honest. i always accepted it as true
restive river
if you prove sqrt(2) + sqrt(3) and sqrt(2) - sqrt(3) are irrational with euclid's lemma, I think the same proof will work in general
like for general a and b, not just 2 and 3
high nebula
hmm
restive river
you could skip the 2 and 3 case if you want, I just thought it might be easier to see if you do that first
high nebula
can't believe i'm having trouble even with that lol
how does the lemma apply to this
restive river
do you have anything so far?
high nebula
tried saying sqrt(2) + sqrt(3) = p/q then squared both sides
restive river
sounds like a good idea
high nebula
but then p²/q² = 5 + 2sqrt(6)
restive river
anything after that?
high nebula
not really. what can i do with that? i mean, it says that sqrt(6) is a rational number and i can't just say that's false without proving sqrt(6) is irrational
restive river
yea, so we should prove sqrt(6) is irrational 🙂
high nebula
huh, alright
i managed to prove that the sqrt(6) is irrational, so i guess i proved that sqrt(2) + sqrt(3) is also irrational
high nebula
but for the general case a and b, i have to show that sqrt(ab) is irrational
high nebula
how'd you prove sqrt(6) is irrational?
proof by contradiction. let sqrt(6) be rational, so there exists coprimes x and y that satisfy sqrt(6) = x/y.
6 = x²/y²
x² = 6y²
this shows that x is even. x = 2n
(2n)² = 6y²
4n² = 6y²
2n² = 3y²
this shows that y is even. this is a contradiction that x and y are coprimes. therefore, sqrt(6) is irrational
wicked oasis
but for the general case a and b, i have to show that sqrt(ab) is irrational
You can save yourself
Take sqrt(a) + sqrt(b) = p/q
sqrt(a) = p/q - sqrt(b)
Now you just have to prove that sqrt(b) is irrational
After squaring both sides
Can save you a little trouble
high nebula
Now you just have to prove that sqrt(b) is irrational
sqrt(b) is irrational by definition
and so is sqrt(a)
restive river
proof by contradiction. let sqrt(6) be rational, so there exists coprimes x and ...
yea ok, that sounds good
wicked oasis
sqrt(b) is irrational by definition
Oh wait just saw the question
Nvm
high nebula
how so
restive river
I just thought of a counterexample 😛
high nebula
which is?
restive river
this is harder to generalize than I originally thought, but I think it'll still work
high nebula
it's easy to show that even numbers not divisible by 4 are irrational when taken the square root. but what about even numbers that are not perfect squares?
like for example the square root of 3 or 5
restive river
ok in that case it would suffice to show sqrt(15) is irrational, right?
after some expanding
high nebula
oh yeah, if you show that sqrt(15) is irrational, then sqrt(3) + sqrt(5) is irrational
surprisingly, it also proves that sqrt(3) - sqrt(5) is irrational
restive river
if sqrt(15) is rational then p^2*15 = q^2 for some relatively prime positive integers p and q. Since 3 divides q*q, 3 divides q by euclid's lemma. Also, 5 divides q for the same reason
so q = 5*3*t for some integer t
and q^2 = 5^2*3^2*t^2
then we can cancel 15's and see 3 and 5 divide p
so p and q are not relatively prime
high nebula
so what you're saying is, if sqrt(ab) is rational, then ab * p² = q² for some integers p and q. since ab divides q*q, a and b divides q. so q = ab * t for some integer t.
and q² = a²b²t²
so ab * p² = a²b²t² => p² = ab * t², which means that p and q are not relatively prime?
restive river
sure, but before anything is said, a and b don't even need to be coprime
high nebula
what do they need to be? and where in my statement do you have to fix it?
restive river
the problem says a and b are not equal
so they just need to not be equal 😛
as well as the other stuff the problem says
high nebula
so what you're saying is, if sqrt(ab) is rational, then ab * p² = q² for some in...
yes but ignoring the original problem, what makes this inherently fail?
it's obvious that if a = b then sqrt(ab) = a
restive river
hmm maybe it does work, let me reread a few times
high nebula
(it doesn't)
restive river
since ab divides q*q, a and b divides q. so q = ab * t for some integer t.
high nebula
i can find a counterexample to it. a = 3 and b = 12
it doesn't necessarily say anywhere that a and b have to be prime or coprime
restive river
i can find a counterexample to it. a = 3 and b = 12
counterexample to what?
oh
yea, that's what I was trying to get at actually
high nebula
to the "proof" i gave
like. what i wrote seems okay but for some reason it's wrong. somehow you have to use the fact that a and b are coprime
restive river
since ab divides q*q, a and b divides q. so q = ab * t for some integer t.
yea, it's this that doesn't work in general
high nebula
hmm
restive river
like. what i wrote seems okay but for some reason it's wrong. somehow you have t...
we don't need a and b to be coprime
restive river
well, sqrt(3) + sqrt(12) is irrational
restive river
so what you're saying is, if sqrt(ab) is rational, then ab * p² = q² for some in...
but we want to do something before going down this route
high nebula
well, sqrt(3) + sqrt(12) is irrational
then there must be a way to prove it without using the fact that sqrt(ab) is irrational, because sqrt(3*12) is not irrational
high nebula
how are we supposed to know if there are factors of a or b that can be taken out of the square root?
restive river
then sqrt(3) + sqrt(12) is just 3sqrt(3)
high nebula
hmm
restive river
how are we supposed to know if there are factors of a or b that can be taken out...
we just suppose there are and deal with them in general
if there aren't, then we're done
high nebula
that makes sense if b is a multiple of a and this multiple is a perfect square. what if it isn't? for example, a = 2 and b = 6
restive river
for any positive integer m>1 which is not a perfect square, sqrt(m) can be written as n*sqrt(p_1*p_2*...*p_k) where b is a positive integer and the p's are prime
high nebula
how is it a multiple of b
high nebula
so let's call them n and m
restive river
ok
high nebula
for simplicity
restive river
I'll edit it
high nebula
hmm that seems reasonable
restive river
do you see why?
high nebula
yes, i do see why
restive river
ok, nice
high nebula
a number is made up of prime numbers
and if it's a prime power, all of its power will go outside the square root unless the last one if it's an odd power
restive river
yep
and there will be some prime that has an odd power since m is not a perfect square
high nebula
hmm, you have that sqrt(a) and sqrt(b) is the same as n * sqrt(a1 * a2 * ... * ak) + m * sqrt(b1 * b2 * ... * bt)
restive river
yea, for some primes a_1, ... a_k, b_1, ..., b_t
btw, can we use the fact that square roots of not perfect squares are irrational?
if so, we don't need to go through the euclid's lemma stuff
if not, it might be easier to prove that as a lemma
high nebula
yea, for some primes a_1, ... a_k, b_1, ..., b_t
they don't necessarily have the same amount of primes
restive river
hmm, you have that sqrt(a) and sqrt(b) is the same as n * sqrt(a1 * a2 * ... * a...
sorry, typo
high nebula
btw, can we use the fact that square roots of not perfect squares are irrational...
that would make it a lot easier if we could. we can try to prove it first
restive river
hmm, you have that sqrt(a) and sqrt(b) is the same as n * sqrt(a1 * a2 * ... * a...
yea, so if we can prove that first - from here we'd have sqrt(a1 * a2 * ... * ak * b1 * b2 * ... * bt) is rational, but that is not the case because a1 * a2 * ... * ak * b1 * b2 * ... * bt is a product of primes with exponents 1 or 2, and at least one of them has an exponent of 1 because a is not equal to b
high nebula
hmmmm
restive river
since a_1, ... a_k are all distinct, and b_1, ..., b_t are all distinct
and from the fact that a positive integer is a perfect square iff every exponent in its prime factorization is even
high nebula
you can have the case where all a1 * a2 * ... * ak is equal to b1 * b2 * ... * bt but with n different than m
like for example a = 2 and b = 8
but then, of course, you can just factor the square root
and since the numbers outside the square root are all integers, it'll just be a multiple of the square root
technically you'd have to prove that an integer multiple of an irrational square root is still irrational but we'll just accept it's true
restive river
yea, if a1 * a2 * ... * ak = b1 * b2 * ... * bt then n * sqrt(a1 * a2 * ... * ak) + m * sqrt(b1 * b2 * ... * bt) = n * m * sqrt(a1 * a2 * ... * ak) is irrational
restive river
technically you'd have to prove that an integer multiple of an irrational square...
that's simple to prove, enough that it might not even need proof
high nebula
yeah, like, if a sqrt(b) is rational but sqrt(b) is irrational, then a sqrt(b) = p/q and then sqrt(b) = p/aq
restive river
yep!
restive river
I believe so!
high nebula
that's honestly so cool
restive river
haha I agree
high nebula
what if we give it more degrees of freedom
native plover
sum1 help me out too 💀
high nebula
there are other help channels
restive river
sum1 help me out too 💀
hey there - try opening a channel 🙂
native plover
there are other help channels
but no helpers
native plover
hey there - try opening a channel 🙂
high nebula
damn
native plover
:c
restive river
what if we give it more degrees of freedom
let me see what the internet has to say
I'd think it's still true though
high nebula
by more degrees of freedom, i mean like, restricting the statement less to make it more useful, yet harder to prove
like for example
restive river
like increasing the amount of square roots of integers summed up?
high nebula
that's also an idea
i was also thinking of this
instead of it being $\sqrt{a}, \sqrt{b} \not\in \mathbb{Z}; \sqrt{a} \pm \sqrt{b} \in \mathbb{R} - \mathbb{Q}$, we can try to see if it works for $n,m \in \mathbb{N}; \sqrt[n]{a}, \sqrt[m]{b} \not\in \mathbb{Z}; \sqrt[n]{a} \pm \sqrt[m]{b} \in \mathbb{R} - \mathbb{Q}$
woven radishBOT
Taiga
restive river
ahhh
high nebula
it sounds a lot harder now, doesn't it?
because now the exponents are not the same
can also go slower with n = m first
i find math beautiful because of these kind of things. one day you're curious, make a hypothesis, prove whether a hypothesis is true, then add or remove degrees of freedom to push it to its limits
well, i'll leave this help channel be for now. this was fun to discuss about
thank you for giving me more insight, yume
restive river
yep, it was fun, thanks 😊
high nebula
.close
merry seal
I don't understand what's going on here
devout snowBOT
I don't understand what's going on here
merry seal
the (1) in first image refers to the second image
this is (2)
oh oh wait I got it
it just switched t with s, and used t as upper bound
smh why is this book like this
.close
devout snowBOT
devout snowBOT
Available help channel!
devout snowBOT
magic rapids
Need help with this please
tawdry trellis
Need help with this please
Raise x + 1/x to the 13th power
magic rapids
okay done it
tawdry trellis
Huh
stuck field
Lol
magic rapids
and?
tawdry trellis
Expand it…
There’s a similar problem involving x^3 + 1/x^3, this one’s probably reasonably similar
magic rapids
tawdry trellis
tawdry trellis
First of all (x + 1/x)^13 not ((x^2 + 1)/x)^13
wooden zodiac
magic rapids
and im stuck
tawdry trellis
Second of all expand it
magic rapids
okay but why is that
tawdry trellis
I don’t understand what’s confusing about that
I’m not saying it’s wrong
I’m saying it’s a bad way to write it
tawdry trellis
okay but why is that
Because you’ll have terms and x^13, 1/x^13
magic rapids
where does the ^2 GO?
magic rapids
go sorry
tawdry trellis
(x^2 + 1)/x = x + 1/x
Just expand (x + 1/x)^13
You did a weird rearrangement that makes the problem harder
restive river
thought you were tryna do (x+1)/x
tawdry trellis
thought you were tryna do (x+1)/x
?
stuck field
Yeah mate? It's already getting confusing here. Don't confuse again lmao
restive river
are you unable to read?
tawdry trellis
I’m wondering what on earth you meant by that statement
restive river
sounds like a you problem
tawdry trellis
Uh huh
stuck field
No, I hardly think your statement made any sense to anyone.
magic rapids
how do i expand that
restive river
💀
tawdry trellis
Do you know the binomial theorem
tawdry trellis
Well what country are you from
magic rapids
hungary
stuck field
Binomiális tétel
magic rapids
yes i got that
stuck field
I don't think so.
If you are given that, there should be some way they expected you to solve this.
I believe
Figure x^2+1/x^2 then x^4+1/x^4 and so on.
But it gets lengthy.
magic rapids
its a shitty online thing for the contry and math students needs to solve a couple for points
there are some hard ones
tawdry trellis
Right well if you don’t know the binomial theorem this’ll be rough