I thought my comments sections were full of _glorious friends of wisdom?_ No solution to either problem yet? Let's see some absolutely aesthetic spellweaving fit for the gods.
I did it this way : 12^b = 2^b * 6^b = 6^a Divide both side by 6^a 2^(a) * 6^(b-a) = 1 2^(b) * 3^(b-a) = 2^0 * 3^0 b = 0 and b = a => b=a=0 But b and a supposed to be natural number, which is a, b > 1 But here it's shown that b < 1 Which contradicts the earlier statement
Awesome video with funny/yet as always cool thumbnail, nice explanations, cool editing and wholesome comment section. This channel is quite underrated.
@@nuklearboysymbiote Two reasons that come to mind: 1) It's a very niche kind of result. 2.) It is the answer to one of Hilbert's Problems (it's hard).
For exercise 2 Hilbert’s 7th states that a^b is transcendental if a is algebraic and b is irrational algebraic. Assume for a contradiction that logm(n) is algebraic. Consider m^(logm(n)) clearly this equals n. m is a natural so is a root of x-m=0 so is algebraic. Logm(n) is algebraic and irrational so Hilbert’s 7th applies and implies that n is transcendental. This is clearly not true therefore logm(n) is not algebraic, therefore is transcendental.
(My scratchwork for in-video exercise at 3:13) First note that: log_6(12) = log_6(6*2) = log_6(6) + log_6(2) = 1 + log_6(2) So we really only need to examine log_6(2) = ln(6)/ln(2). Set this equal to the ratio of natural numbers a and b: ln6/ln2 == a/b b ln6 == a ln2 ln(6^b) == ln(2^a) ln(2^b*3^b) == ln(2^a) 2^b*3^b == 2^a 3^b == 2^(a-b) The conclusion to the proof-by-contradiction is the same as in the first question: Since b and (a-b) are both integers, we are trying to set an odd number -- 3^b -- equal to an even number -- 2^(a-b), which is not possible.
for the one at 3:17 I rewrote log_6(12) as 1 + log_6(2), the reason we can assume a,b \in \mathbb{N} is because ln has no local extrema on \mathbb{R}^+ by the derivative test from real analysis so that log_6(2) > 0. Then as you did before change of base gives bln2 = aln6 so ln2^b = ln6^a s.t. 2^b = 6^a, which is a contradiction because 3 divides the RHS and not the LHS. (log_6(12) is then irrational because 1 + [irrational] = irrational, to see this use an RAA argument)
3:20 I solved it like this: log6(12)= a/b implies a/b= ln(12)/ln(6) = ln(3×2×2)/ln(3×2) = (ln 3 + ln 2 + ln 2)/(ln 3 + ln 2) = 1 + ln 2 / ln 3 = a/b But we just proved ln 3 / ln 2 is transcendental a moment ago, thus so is ln 2 / ln 3. Then 1 + ln 2 / ln 3 is transcendental as well, and so can't be written as a/b.
3:20 first of all, you can change this expression like you did with log2(3). Then you get 12^b=6^a which is the same as 6^b * 2^b=6^a. If a and b are natural numbers then this gives us a contradiction because a power of 6 multiplied by a power if 2 will never be equal to some power of 6.
Every time u raise 12 to a power the prime factorization changes by adding two 2s and one 3, whereas with 6, u add one 2 and one 3. Therefore, in an exponent of 12, the ratio of 2s to 3s in always 2. But in an exponent of 6, the ratio of 2s to 3s in its prime factorization must be 1. And 2 is distinct from 1. QED
Suppose log base 6 of 17 = a/b with a and b being natural numbers, then you have natural log of 6/ natural log of 17 = a/b and the last passage is 6^a = 17^b
You could do it without the extra step. Given log2(3)=a/b, raise both sides to power of 2 => 3 = 2^(a/b), raise both sides to power of b => 3^b=2^a. This is 2 steps while your way is 3 steps. Also going from bln(3)=aln(2) to 3^b=2^a feels like you are skipping a step.
I like the Vi Hart video where she explained logarithms: th-cam.com/video/N-7tcTIrers/w-d-xo.html . Logarithm is just asking the following question: how many '*a' steps does it take to reach the number b? So log2(4)=2, because to reach 4 you need to take two '*2' steps. Log2(√2)=1/2, because √2 is a halfway towards 2 (you need to take two '*√2' steps to reach 2). Log2(1/2)=-1, because to reach 1/2, you need to divide by 2 once (essentially, this is taking a '*2' step back). To which I added (but unfortunately comments are no longer available at her videos): That's all nice, until somebody asks: how many '*2' steps does it take to reach the number 3? Is it one step? No, that's not enough. Two steps? Too much. One and a half? Still not it. It can be seen that no rational number of steps will do, because no power of 2 is a power of 3.
How do you know that no rational number satisfies that (eg. 1.58)? Judging from how you "prove" it, it seems like you were just guessing and at best just making a conjecture.
@@FrostDirt That log2(3) is not a rational number means *precisely* that no power of 2 is a power of 3. Consider for contradiction: * Suppose log2(3)=p/q (where p and q are positive integers). * Exponentiate both sides to the power of 2. We get: 3=2^(p/q). * Take q-th power of both sides. We get: 3^q=2^p. * But as I have said, no power of 2 is a power of 3. (This follows from the fact that every positive integer has a unique factoring into primes - or, as the video says, that an odd number can't be equal to an even number.) QED. That each step is valid follows from the fact that functions x->c^x and x->x^c, for c being a constant greater than 1, are both strictly increasing (and so injective) on positive real numbers. We also need the theorem that (a^b)^c=a^(b*c). Of course, we also need to know that log2(3) isn't an integer, but this is obvious: 2^1=2, and 2^2=4, so the logarithm must be strictly between 1 and 2.
@@MikeRosoftJH That's not what he's saying. It's quite clear that log2(3) is irrational if and only if no power of 2 is equal to a power of 3. He's asking where the assertion that either of those two things is true is coming from. (Powers of 2 are even, powers of 3 are odd. That's what I think he's looking for.)
@@EpicMathTime I'm not sure, but have I missed something in the above post? I think that in the post above I have shown that: * The fact that no power of 2 is a power of 3 implies that log2(3) is irrational; and * It's indeed the case that no power of 2 is a power of 3 (because 1) an odd number can't be equal to an even number, and 2) every positive integer has a unique factorization into primes). Of course, by "no power of 2" (or 3) I mean a positive integer power.
Honestly, I would have preferred if I said that. I generally don't consider 0 to be a natural number, but it's not a conscious decision I made or anything like that. I just kind of... "don't", but sometimes I forget I don't. (Probably because it's not a conscious decision.)
Actually, a = b = 0 is a natural solution to 2b = a & b = a, because 0 is a natural number. However, it was stated previously in the exercise that it was already known that a > 0 & b > 0 since 12 > 6, so this is not a problem, and the argument presented in the video is still rigorous and valid.
log(6, 12) = log(6, 6) + log(6, 2) = 1 + 1/log(2, 6) = 1 + 1/[log(2, 2) + log(2, 3)] = 1 + 1/[1 + log(2, 3)] is not rational, because 1 + 1/[1 + log(2, 3)] = a/b iff 1/[1 + log(2, 3) = a/b - 1 = (a - b)/b iff 1 + log(2, 3) = b/(a - b) iff log(2, 3) = b/(a - b) - 1 = (2b - a)/(a - b) = a'/b', which contradicts the previous proof. If n = 0, then log(m, 0) = 0, which is rational for all m > 1 in N\{0}. If n > 0, then log(m, n) > 0, for all m > 1, with n & m elements of N\{0}. Therefore, log(m, n) is rational iff log(m, n) = a/b for some a & b elements of N\{0}. log(m, n) = a/b iff n^b = m^a. By the fundamental theorem of arithmetic, every nonzero natural number has a unique prime number factorization. Therefore, p divides n^b iff p divides m^a. If p divides n^b, then it must divide n, and if it also divides m^a, then it must divide m as well. Therefore, p divides n iff p divides m. Since this holds for any prime p, it follows that n and m have the same set of prime factors. Call this P. Therefore, let n = Π{|P| + 1 > k > 0; p(k)^s(k)}, and m = Π{|P| + 1 > k > 0; p(k)^t(k)}, and m^a = n^b iff a·t(k) = b·s(k) for every natural |P| + 1 > k > 0. Since a > 0, b > 0, t(k) > 0, and s(k) > 0, a·t(k) = b·s(k) iff s(k)/t(k) = a/b for all given k. Meanwhile, s(k)/t(k) = a/b iff n^b = m^a with n & m defined above. n^b = m^a iff b·ln(n) = a·ln(m) iff ln(n)/ln(m) = a/b = log(m, n). If log(m, n) is not rational, then it transcendental by the Gelfond-Schneider theorem.
I thought my comments sections were full of _glorious friends of wisdom?_ No solution to either problem yet? Let's see some absolutely aesthetic spellweaving fit for the gods.
I did it this way :
12^b = 2^b * 6^b = 6^a
Divide both side by 6^a
2^(a) * 6^(b-a) = 1
2^(b) * 3^(b-a) = 2^0 * 3^0
b = 0 and b = a => b=a=0
But b and a supposed to be natural number, which is a, b > 1
But here it's shown that b < 1
Which contradicts the earlier statement
it turns out--- EXERCISE FOR THE VIEWER
-Every textbook ever written
2:00 ah yes the fundamental theorem of engineering
e = Pi = 3
@@HugoHabicht12 =sqrt(g)
and pi^2=g
Me doing math: "Hmm, I'm a bit tired of doing math let's go watch youtube to refresh."
*Proceeds to watch epic math*, Nice.
1:52 shows why this man is a true mathematician
Great video! Nice to see you doing so well. ~ Love Andrew’s Mom 💕
Thank you! I hope you guys are doing well!
@@EpicMathTime we are 😊. You should come for Christmas 🎄 & get a pic w Andrew!!
Hah, I don't think I can make it this time, but a meet-up is inevitable in the future!
@@EpicMathTime sound good. Flammable Math should come too! 😊
Awesome video with funny/yet as always cool thumbnail, nice explanations, cool editing and wholesome comment section. This channel is quite underrated.
Gelfond-Schneider Theorem.
Edit: you should explicitly rule out a=b=0, for otherwise m^a=n^b always (using the convention 0^0=1).
Why aren't there youtube videos on that theorem
@@nuklearboysymbiote Two reasons that come to mind:
1) It's a very niche kind of result.
2.) It is the answer to one of Hilbert's Problems (it's hard).
@@EpicMathTime all the more reason for content… i'll look into it
@ゴゴ Joji Joestar ゴゴ ye i couldnt find it on wiki
For exercise 2
Hilbert’s 7th states that a^b is transcendental if a is algebraic and b is irrational algebraic.
Assume for a contradiction that logm(n) is algebraic.
Consider m^(logm(n)) clearly this equals n.
m is a natural so is a root of x-m=0 so is algebraic.
Logm(n) is algebraic and irrational so Hilbert’s 7th applies and implies that n is transcendental. This is clearly not true therefore logm(n) is not algebraic, therefore is transcendental.
This channel has absolutely exploded in quality. Well done.
(My scratchwork for in-video exercise at 3:13) First note that:
log_6(12) = log_6(6*2) = log_6(6) + log_6(2) = 1 + log_6(2)
So we really only need to examine log_6(2) = ln(6)/ln(2). Set this equal to the ratio of natural numbers a and b:
ln6/ln2 == a/b
b ln6 == a ln2
ln(6^b) == ln(2^a)
ln(2^b*3^b) == ln(2^a)
2^b*3^b == 2^a
3^b == 2^(a-b)
The conclusion to the proof-by-contradiction is the same as in the first question: Since b and (a-b) are both integers, we are trying to set an odd number -- 3^b -- equal to an even number -- 2^(a-b), which is not possible.
I like that! It never occured to me to bring it to a similar "even = odd" contradiction.
@@EpicMathTime Thanks!
This man deserves way more subs with this kind of quality of content
Did you srsly just "the proof is left as an exercise for the reader" us?
Oh yes he did... That cheeky bastard. XD
2:10 You know when you're a mathematician, when instead of 'A' you write alpha
Total chad here.
for the one at 3:17 I rewrote log_6(12) as 1 + log_6(2), the reason we can assume a,b \in \mathbb{N} is because ln has no local extrema on \mathbb{R}^+ by the derivative test from real analysis so that log_6(2) > 0. Then as you did before change of base gives bln2 = aln6 so ln2^b = ln6^a s.t. 2^b = 6^a, which is a contradiction because 3 divides the RHS and not the LHS. (log_6(12) is then irrational because 1 + [irrational] = irrational, to see this use an RAA argument)
God damn it I hate how you ended this video but I also love it and can't wait to try when I'm done procrastinating at work by watching math videos.
No one has given write-ups yet, you should honor us with yours!
3:20 I solved it like this:
log6(12)= a/b implies
a/b= ln(12)/ln(6)
= ln(3×2×2)/ln(3×2)
= (ln 3 + ln 2 + ln 2)/(ln 3 + ln 2)
= 1 + ln 2 / ln 3 = a/b
But we just proved ln 3 / ln 2 is transcendental a moment ago, thus so is ln 2 / ln 3.
Then 1 + ln 2 / ln 3 is transcendental as well, and so can't be written as a/b.
*irrational, not transcendental
3:20 first of all, you can change this expression like you did with log2(3). Then you get 12^b=6^a which is the same as 6^b * 2^b=6^a. If a and b are natural numbers then this gives us a contradiction because a power of 6 multiplied by a power if 2 will never be equal to some power of 6.
4:43 why does this follow? why cant the equality arise otherwise?
The fundamental theorem of arithmetic.
Every time u raise 12 to a power the prime factorization changes by adding two 2s and one 3, whereas with 6, u add one 2 and one 3. Therefore, in an exponent of 12, the ratio of 2s to 3s in always 2. But in an exponent of 6, the ratio of 2s to 3s in its prime factorization must be 1. And 2 is distinct from 1. QED
Suppose log base 6 of 17 = a/b with a and b being natural numbers, then you have natural log of 6/ natural log of 17 = a/b and the last passage is 6^a = 17^b
5:49 then n=m^(b/a)
This sounds alot like you're saying a log is rational if it's solution is rational XD
It effectively does say that, that's why it doesn't really offer any insight.
Proof: left as an exercise to the reader.
Me: bruh! >:(
You could do it without the extra step. Given log2(3)=a/b, raise both sides to power of 2 => 3 = 2^(a/b), raise both sides to power of b => 3^b=2^a. This is 2 steps while your way is 3 steps. Also going from bln(3)=aln(2) to 3^b=2^a feels like you are skipping a step.
I like the Vi Hart video where she explained logarithms: th-cam.com/video/N-7tcTIrers/w-d-xo.html . Logarithm is just asking the following question: how many '*a' steps does it take to reach the number b? So log2(4)=2, because to reach 4 you need to take two '*2' steps. Log2(√2)=1/2, because √2 is a halfway towards 2 (you need to take two '*√2' steps to reach 2). Log2(1/2)=-1, because to reach 1/2, you need to divide by 2 once (essentially, this is taking a '*2' step back). To which I added (but unfortunately comments are no longer available at her videos): That's all nice, until somebody asks: how many '*2' steps does it take to reach the number 3? Is it one step? No, that's not enough. Two steps? Too much. One and a half? Still not it. It can be seen that no rational number of steps will do, because no power of 2 is a power of 3.
How do you know that no rational number satisfies that (eg. 1.58)? Judging from how you "prove" it, it seems like you were just guessing and at best just making a conjecture.
@@FrostDirt That log2(3) is not a rational number means *precisely* that no power of 2 is a power of 3. Consider for contradiction:
* Suppose log2(3)=p/q (where p and q are positive integers).
* Exponentiate both sides to the power of 2. We get: 3=2^(p/q).
* Take q-th power of both sides. We get: 3^q=2^p.
* But as I have said, no power of 2 is a power of 3. (This follows from the fact that every positive integer has a unique factoring into primes - or, as the video says, that an odd number can't be equal to an even number.) QED.
That each step is valid follows from the fact that functions x->c^x and x->x^c, for c being a constant greater than 1, are both strictly increasing (and so injective) on positive real numbers. We also need the theorem that (a^b)^c=a^(b*c). Of course, we also need to know that log2(3) isn't an integer, but this is obvious: 2^1=2, and 2^2=4, so the logarithm must be strictly between 1 and 2.
@@MikeRosoftJH That's not what he's saying. It's quite clear that log2(3) is irrational if and only if no power of 2 is equal to a power of 3. He's asking where the assertion that either of those two things is true is coming from.
(Powers of 2 are even, powers of 3 are odd. That's what I think he's looking for.)
@@EpicMathTime I'm not sure, but have I missed something in the above post? I think that in the post above I have shown that:
* The fact that no power of 2 is a power of 3 implies that log2(3) is irrational; and
* It's indeed the case that no power of 2 is a power of 3 (because 1) an odd number can't be equal to an even number, and 2) every positive integer has a unique factorization into primes).
Of course, by "no power of 2" (or 3) I mean a positive integer power.
I’ve an image of the answer to proving log_6(12) is irrational.now how to post it here
Ok never mind , I have it right here:
drive.google.com/file/d/1SqqeE47Xaiu97OauocotbF5fXWb5XDWN/view?usp=drivesdk.what do you think did I do it right?
Just realized I forgot link sharing so here you go again : drive.google.com/file/d/1SqqeE47Xaiu97OauocotbF5fXWb5XDWN/view?usp=drivesdk
Wow this video was good
5:10 2b =a ,b=a doesnt have a solution if a,b are natural. You get the same result, but YEP.
Honestly, I would have preferred if I said that. I generally don't consider 0 to be a natural number, but it's not a conscious decision I made or anything like that. I just kind of... "don't", but sometimes I forget I don't. (Probably because it's not a conscious decision.)
Actually, a = b = 0 is a natural solution to 2b = a & b = a, because 0 is a natural number. However, it was stated previously in the exercise that it was already known that a > 0 & b > 0 since 12 > 6, so this is not a problem, and the argument presented in the video is still rigorous and valid.
That guy is kewl, wish people who taught maths, science etc in india also looked this kewl
Where's A Good Place To Stop when you need them.
truly epic
Seventieth!
How the fuck he writes facing the camera in these videos is a more important problem than the Riemann hypothesis
Ok here's my contradiction
log_6 (12) = a/b
ln(12) / ln(6) = a/b
b * ln(12) = a * ln(6)
12^b = 6^a
(2)^b * (6)^b = 6^a
(2)^2b * 3^b = 2^a * 3^a
2^(2b - a) = 3^(a-b)
Both 2b-a and a-b are natural numbers, therefore 2^(2b-a) cannot equal 3^(a-b)
This is fucking nuts
edit: I love it
log(6, 12) = log(6, 6) + log(6, 2) = 1 + 1/log(2, 6) = 1 + 1/[log(2, 2) + log(2, 3)] = 1 + 1/[1 + log(2, 3)] is not rational, because 1 + 1/[1 + log(2, 3)] = a/b iff 1/[1 + log(2, 3) = a/b - 1 = (a - b)/b iff 1 + log(2, 3) = b/(a - b) iff log(2, 3) = b/(a - b) - 1 = (2b - a)/(a - b) = a'/b', which contradicts the previous proof.
If n = 0, then log(m, 0) = 0, which is rational for all m > 1 in N\{0}. If n > 0, then log(m, n) > 0, for all m > 1, with n & m elements of N\{0}. Therefore, log(m, n) is rational iff log(m, n) = a/b for some a & b elements of N\{0}. log(m, n) = a/b iff n^b = m^a. By the fundamental theorem of arithmetic, every nonzero natural number has a unique prime number factorization. Therefore, p divides n^b iff p divides m^a. If p divides n^b, then it must divide n, and if it also divides m^a, then it must divide m as well. Therefore, p divides n iff p divides m. Since this holds for any prime p, it follows that n and m have the same set of prime factors. Call this P. Therefore, let n = Π{|P| + 1 > k > 0; p(k)^s(k)}, and m = Π{|P| + 1 > k > 0; p(k)^t(k)}, and m^a = n^b iff a·t(k) = b·s(k) for every natural |P| + 1 > k > 0. Since a > 0, b > 0, t(k) > 0, and s(k) > 0, a·t(k) = b·s(k) iff s(k)/t(k) = a/b for all given k. Meanwhile, s(k)/t(k) = a/b iff n^b = m^a with n & m defined above. n^b = m^a iff b·ln(n) = a·ln(m) iff ln(n)/ln(m) = a/b = log(m, n).
If log(m, n) is not rational, then it transcendental by the Gelfond-Schneider theorem.
Fourth ✌🏼
blueballed
I like the new haircut :D
When Epic Math Time finally uploads:
Long have I waited...
Second!!
FIRSTTTT
fifth