If we had that m/n isn't integer, we would have that m is less than 2n. This means that the exponent of n is less than the exponent of m, which is a contradiction since m>n.
(m/n)=n^[(m/n)-2] 1. If (m/n)2, then the right hand side is a positive rational power of an integer. Positive rational powers of integers are either irrational or integers, so the right hand side is either an integer or irrational. However, the left hand side is rational. The only way they can be equal is if they are equal to an integer.
(m/n)=n^[(m/n)-2] 1. If (m/n)2, then the right hand side is a positive rational power of an integer. Positive rational powers of integers are either irrational or integers, so the right hand side is either an integer or irrational. However, the left hand side is rational. The only way they can be equal is if they are equal to an integer.
@@Keithfert490 You are right with "either irrational or integer", but that's a whole theorem. At least, he should say, it's "trivial". But he didn't mention this fact at all!
@rainerzufall42 but you have to prove it, exactly for the same reason you have to prove that x has to be an integer. Of course you answer me "no" giving me a proof, as well as I can say "no" to your initial question about x being or not a non-integer between 2 and 4, after giving a proof of it. The point of my (and your) observation is not IF x can be a rational non-integer, but why Penn didn't face that case.
@ In my original post, I said 2 to 4 for a reason. I did not prove, why it should not be between 1 and 2 (x - 2 < 0), and I did not prove, why it shoud not be between 4 and 5 (proved it later). I just took the fact that for x = 4: x = 2^(x - 2) is equal and for greater values, the power dominates, so I chose 2 to 4 as my interval. That was a choice, not a theorem, so I didn't have to prove it. All under the premise, that it could be possible (which it isn't) that x is a non-integer rational!
Even Peano originally thought of 1 as the smallest natural number, the wikipedia citation for this is "1889 in Arithmetices Principia. p. 1." However he later changed it to 0. If you think about it 0 is not natural at all, you don't ever use it for counting (which is what natural numbers are all about)
7:00 well, we cannot use induction for a non-natural variable, though the point is pretty clear. Comparing derivatives could be a simple approach. m/n having to be integer is also obvious, yet hasn't been stated beforehand.
By the fundamental theorem of arithmetic, we can assume that m=n^k. So we can rewrite the problem n^(kn)=n^(n^k-n). Taking the logarithm of both sides, we get n=1 or kn = n^k-n. From the second equation, we get k+1 = n^(k-1). If n >= 4, then k+1 >= 4^(k-1) but it's impossible when k > 1, and when k = 1, 2 ≠ 1, which means n
I came here to say this. I did something slightly different, I solved for n = (k+1)^(1/(k-1)) and observed that the function on the RHS is decreasing so there are only a small number of values of k to check. I then checked values of k rather than n, but the result was the same.
I disagree with your assumption of use of the theorem. You are quickly implying that m/n is an integer without plea or sake of an easy proof. But still on still your solution is great!
No, you cannot assume that m=n^k without proper justification Because considering an equation m^a = n^b, where a, b are some functions of n, m(a, and b are integers of course) does not immediately imply that m=n^k nor that n=m^k. Example: a(n, m) = m-n-1, b(n, m) = n-2, the equation n^a = m^b has a solution n=4, m=8, both of which are not integer powers of each other
My solution involved more cases: Case 1: Assume m = n. Then m^n = n^0 = 1 ==> m = n = 1. Furthermore, only assuming m = 1 implies n = 1 and only assuming n = 1 implies m = 1. So for all future cases, we can assume both m and n are at least 2. Case 2: Assume n > m. Then the RHS is a natural number bigger than 1 raised to a negative exponent i.e. not a natural number. But the LHS is a natural number so no sol. Case 3: Assume n < m. In other words, we can set m = n + k for some natural number k. So our equation now becomes (n+k)^n = n^k. This further breaks into 2 cases: Case3a: Assume k = n^n + k^n >= n^k + k^n > n^k so no sol. Case 3b: Assume k > n. In other words, we can set k = n + j for some natural number j. So our equation now becomes (2n+j)^n = n^(n+j). Dividing both sides by n^n gives us (2 + (j/n))^n = n^j. The LHS can only be a natural number if j is a multiple of n. From there, it is helpful to simply start enumerating multiples of n: Let j = n ==> 3^n = n^n. Take both sides to the 1/n power ==> n = 3. From there, the original equation becomes 27m^3 = 3^m. It's not too hard to find that m = 9 is a solution. You can use calculus or maybe induction to show that is the only solution. Let j = 2n ==> 4^n = n^(2n). Take both sides to the 1/n power ==> 4 = n^2 ==> n = 2. From there, the original equation becomes 4m^2 = 2^m. It's not too hard to find that m = 8. Similar to above, there is only 1 solution. Let j = 3n ==> 5^n = n^(3n). Take both sides to the 1/n power ==> 5 = n^3 ==> no solution. From here, you run into the same problem as j gets ever bigger. Namely that the LHS grows far too slowly to keep up with the RHS and would require n to take on some value between 1 and 2, which isn't a natural number.
Quite easy to show that m/n is indeed an integer: m/n = n^(m/n-2). The exponent in the right hand side is either rational, or an integer. A natural number n raised to q = m/n-2 power is either irrational, rational or an integer. In case it's an integer - we get that m/n = k, or m = n*k, so m is multiple of n, so m/n-2 must be an integer. In case it's irrational - the equation cannot hold (since on LHS we have a rational) In case it's rational (like 25^(-1/2) = 1/5) we only have the possibility of the RHS being an 1/K, where K is some integer. And because of that, m/n = 1/K => n = K*m, which cannot hold since m>n. So m/n must be an integer.
From the fundamental theorem of arithmetic, m^n=r^s implies m=a^{s'} and r=a^{n'} where s'=s/gcd(n,s), n'=n/gcd(n,s) and some natural number a. Substituting these for m and n then equating the exponents, we can solve the problem.
(m/n)=n^[(m/n)-2] 1. If (m/n)2, then the right hand side is a positive rational power of an integer. Positive rational powers of integers are either irrational or integers, so the right hand side is either an integer or irrational. However, the left hand side is rational. The only way they can be equal is if they are equal to an integer.
Here's a very simple explanation of why X=m/n can't be a non-integer. Consider the equation X=n^(x-2), equation (*). Suppose X is not an integer and suppose X=p/q, where p and q are relatively prime (it could be that (m, n)=(p, q), but not necessarily). There are 2 cases: 1) the q-th root of n is an integer (so n^X is an integer). Then the LHS of equation (*) is non-integral, while the RHS is integral. 2) the q-th root of n is not an integer. It can't be a rational number, because the base n is an integer, therefore it has to be irrational. So then the RHS of (*) is irrational and the LHS is rational. Contradiction in both cases. Hope this helps :)
This is brilliant ❤️ You explained this like a computer scientist. The prove by contradiction had an underlying assumption to prove against and had a pair of sets of complementary arguments that where contradictory. So the assumption had to be overturned
At 4.10 the inequality you are writting could be false if m < 2n (the exponent would be negative). You are also assuming x is an integer. this is clear if m>= 2n as x = n^{m/n - 2}, nut not if m
Hey! Some years ago I stated a somewhat similar problem: Find all triples m,n,k st mⁿ = (m+k)^(n-k), k≠0 The simplest is 2,4,2 but i recall finding an infinite set of solutions. Is was 2018 and I offered a melon as a prize 😂
If you're a "0^0=1" chad it works just fine apart from the arbitrary "solutions must be natural" restriction. If your a "0^0 is undefined" virgin then you can't have that solution.
@@rukanee2883A lot of subfields (the majority of them, IIRC) define 0^0 to be 1, as it aligns well with lots of things, doesn't seem to cause any contradictions, and makes exponentiation total over the Natural Numbers including 0.
@@rukanee2883 m & n being natural does not prevent m=n. Indeed, n^(m-n) being natural was used to justify m>=n. Thus, as stated, m and m^n being natural doesn't imply that n^(m-n) is natural.
Wdym part from? 1^2 = 1 Also not exactly a coincidence. We’re looking at an equation where natural number variables satisfy an exponential equation. While it can’t be guaranteed by this reasoning, it shouldn’t be surprising to find that one solution is a natural exponent of the other. Good on you for asking questions, though! That’s always important in mathematics.
Lowercase m,n is an unfortune choice of letters, especially with your handwriting, Michael. The sound is also similar. Perhaps it is just me, English is not my native language.
It’s just the standard convention where when you are working with Natural numbers your arbitrary variables are typically n and m. (And then if you are summing or multiplying over a range of Naturals then the variable used in the sum/product is k or j ranging from, say, 1 to n.). So while m and n aren’t necessarily actually the best choice of letters in terms of clarity depending on handwriting and sounding similar when spoken, it’s just kind of tradition that people seem to use them. 🤷♂️
9:50 Zero anyway is not a solution, because plugging 0 at the original equation you either get 0=1, which is not true, or 0^0=0^0 which is also not true (since 0^0 cannot be evaluated)
Isn't m=n=0 also a valid solution? 0^0 = 0^(0-0) = 1. Edit: For anyone replying to this with "0^0 is undefined/indefinite/etc.": It is only indefinite/undefined/whatever if you're talking about the limit of two functions, f(x)^g(x), where the limit as x->c is zero for both. Evaluated as a _number,_ 0^0 = 1. It has to. Otherwise, e^x is not equal to its Taylor series expansion at 0.
@@forcelifeforce That is the only value that it can consistently have, and in set theory and integer exponentiation--which is what matters in this context--it _is_ universally agreed to be 1. No other value matters.
@@TomFarrell-p9z I always forget that different groups define it differently. The way I was taught, 0 is a natural number. It seems Professor Penn falls in the camp that excludes 0.
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5:46 why does x have to be integer?
I also didn't get it. Why not check 37/13?
If we had that m/n isn't integer, we would have that m is less than 2n. This means that the exponent of n is less than the exponent of m, which is a contradiction since m>n.
(m/n)=n^[(m/n)-2]
1. If (m/n)2, then the right hand side is a positive rational power of an integer. Positive rational powers of integers are either irrational or integers, so the right hand side is either an integer or irrational. However, the left hand side is rational. The only way they can be equal is if they are equal to an integer.
@@alexeykamenev1348 How can any power of 37 and 13 be equal?
@nicolacircella5259 If m/n=37/13 then m>2n so I don't know what your point is supposed to be
Why does m/n have to belong to the set {2,3,4}? Isn't m/n a (positive) rational > 1, not a natural number?
(m/n)=n^[(m/n)-2]
1. If (m/n)2, then the right hand side is a positive rational power of an integer. Positive rational powers of integers are either irrational or integers, so the right hand side is either an integer or irrational. However, the left hand side is rational. The only way they can be equal is if they are equal to an integer.
@@Keithfert490 Thank you for the careful explanation.
@@Keithfert490 You are right with "either irrational or integer", but that's a whole theorem. At least, he should say, it's "trivial". But he didn't mention this fact at all!
@rainerzufall42 agreed that he should've addressed it for sure!
My question exactly
8:50 Not a word about why x can't be a non-integer between 2 and 4 ?
It could also be a rational non-integer between 4 and 5.
@@MarcoMate87 No, for n > 1, we have n^(x-2) > x, if x > 4. Best case for x = 4 is 2^(x-2) = 4 = x (n = 2).
@@MarcoMate87 Apart from that, x has to be integer, he just doesn't mention why!
@rainerzufall42 but you have to prove it, exactly for the same reason you have to prove that x has to be an integer. Of course you answer me "no" giving me a proof, as well as I can say "no" to your initial question about x being or not a non-integer between 2 and 4, after giving a proof of it. The point of my (and your) observation is not IF x can be a rational non-integer, but why Penn didn't face that case.
@ In my original post, I said 2 to 4 for a reason. I did not prove, why it should not be between 1 and 2 (x - 2 < 0), and I did not prove, why it shoud not be between 4 and 5 (proved it later). I just took the fact that for x = 4: x = 2^(x - 2) is equal and for greater values, the power dominates, so I chose 2 to 4 as my interval. That was a choice, not a theorem, so I didn't have to prove it. All under the premise, that it could be possible (which it isn't) that x is a non-integer rational!
@9:50 you exclude 0 as being a member of the set natural numbers. Peano's first axiom states that 0 is a natural number
That's the "us" natural set.
Even Peano originally thought of 1 as the smallest natural number, the wikipedia citation for this is
"1889 in Arithmetices Principia. p. 1." However he later changed it to 0. If you think about it 0 is not natural at all, you don't ever use it for counting (which is what natural numbers are all about)
@@Kettwiesel25 Peano changed his mind. If i think about it 0 is a natural number, and Z-F also agreed with their axioms that zero is a natural number
7:00 well, we cannot use induction for a non-natural variable, though the point is pretty clear. Comparing derivatives could be a simple approach.
m/n having to be integer is also obvious, yet hasn't been stated beforehand.
m/n being integer has to be proven, or at least being talked about...
@rainerzufall42 No, but it is pretty trivial. Threw me for a sec though too when he said "let's use induction"
@ He can do this kind of induction, if and when he proves, that m/n is integer (natural)! Not earlier!
By the fundamental theorem of arithmetic, we can assume that m=n^k. So we can rewrite the problem n^(kn)=n^(n^k-n). Taking the logarithm of both sides, we get n=1 or kn = n^k-n. From the second equation, we get k+1 = n^(k-1). If n >= 4, then k+1 >= 4^(k-1) but it's impossible when k > 1, and when k = 1, 2 ≠ 1, which means n
I came here to say this. I did something slightly different, I solved for n = (k+1)^(1/(k-1)) and observed that the function on the RHS is decreasing so there are only a small number of values of k to check. I then checked values of k rather than n, but the result was the same.
Doesn't your approach assume n is prime? How can that be preemptively true?
I disagree with your assumption of use of the theorem. You are quickly implying that m/n is an integer without plea or sake of an easy proof.
But still on still your solution is great!
No, you cannot assume that m=n^k without proper justification
Because considering an equation m^a = n^b, where a, b are some functions of n, m(a, and b are integers of course) does not immediately imply that m=n^k nor that n=m^k. Example: a(n, m) = m-n-1, b(n, m) = n-2, the equation n^a = m^b has a solution n=4, m=8, both of which are not integer powers of each other
Oops why I couldn't consider the possibility k is a rational number
My solution involved more cases:
Case 1: Assume m = n. Then m^n = n^0 = 1 ==> m = n = 1. Furthermore, only assuming m = 1 implies n = 1 and only assuming n = 1 implies m = 1. So for all future cases, we can assume both m and n are at least 2.
Case 2: Assume n > m. Then the RHS is a natural number bigger than 1 raised to a negative exponent i.e. not a natural number. But the LHS is a natural number so no sol.
Case 3: Assume n < m. In other words, we can set m = n + k for some natural number k. So our equation now becomes (n+k)^n = n^k. This further breaks into 2 cases:
Case3a: Assume k = n^n + k^n >= n^k + k^n > n^k so no sol.
Case 3b: Assume k > n. In other words, we can set k = n + j for some natural number j. So our equation now becomes (2n+j)^n = n^(n+j). Dividing both sides by n^n gives us (2 + (j/n))^n = n^j. The LHS can only be a natural number if j is a multiple of n. From there, it is helpful to simply start enumerating multiples of n:
Let j = n ==> 3^n = n^n. Take both sides to the 1/n power ==> n = 3. From there, the original equation becomes 27m^3 = 3^m. It's not too hard to find that m = 9 is a solution. You can use calculus or maybe induction to show that is the only solution.
Let j = 2n ==> 4^n = n^(2n). Take both sides to the 1/n power ==> 4 = n^2 ==> n = 2. From there, the original equation becomes 4m^2 = 2^m. It's not too hard to find that m = 8. Similar to above, there is only 1 solution.
Let j = 3n ==> 5^n = n^(3n). Take both sides to the 1/n power ==> 5 = n^3 ==> no solution. From here, you run into the same problem as j gets ever bigger. Namely that the LHS grows far too slowly to keep up with the RHS and would require n to take on some value between 1 and 2, which isn't a natural number.
Quite easy to show that m/n is indeed an integer:
m/n = n^(m/n-2). The exponent in the right hand side is either rational, or an integer. A natural number n raised to q = m/n-2 power is either irrational, rational or an integer.
In case it's an integer - we get that m/n = k, or m = n*k, so m is multiple of n, so m/n-2 must be an integer.
In case it's irrational - the equation cannot hold (since on LHS we have a rational)
In case it's rational (like 25^(-1/2) = 1/5) we only have the possibility of the RHS being an 1/K, where K is some integer. And because of that, m/n = 1/K => n = K*m, which cannot hold since m>n.
So m/n must be an integer.
From the fundamental theorem of arithmetic, m^n=r^s implies m=a^{s'} and r=a^{n'} where s'=s/gcd(n,s), n'=n/gcd(n,s) and some natural number a. Substituting these for m and n then equating the exponents, we can solve the problem.
8:30 (approx,) Hold on... Why would m/n (aka x) be an integer ?
Why must $n|m$? Would it be possible for $\frac{m}{n} \in \mathbb{Q}$?
(m/n)=n^[(m/n)-2]
1. If (m/n)2, then the right hand side is a positive rational power of an integer. Positive rational powers of integers are either irrational or integers, so the right hand side is either an integer or irrational. However, the left hand side is rational. The only way they can be equal is if they are equal to an integer.
But what is a bad place to stop?
There, I guess ----> .
(Yes, that was a full-stop)
Here's a very simple explanation of why X=m/n can't be a non-integer.
Consider the equation X=n^(x-2), equation (*). Suppose X is not an integer and suppose X=p/q, where p and q are relatively prime (it could be that (m, n)=(p, q), but not necessarily). There are 2 cases:
1) the q-th root of n is an integer (so n^X is an integer). Then the LHS of equation (*) is non-integral, while the RHS is integral.
2) the q-th root of n is not an integer. It can't be a rational number, because the base n is an integer, therefore it has to be irrational. So then the RHS of (*) is irrational and the LHS is rational.
Contradiction in both cases. Hope this helps :)
This is brilliant ❤️
You explained this like a computer scientist.
The prove by contradiction had an underlying assumption to prove against and had a pair of sets of complementary arguments that where contradictory.
So the assumption had to be overturned
@@wisdomokoro8898 Thanks! Glad it was helpful. I was confised myself in the beginning as well, but this reasoning cleared things out.
At 4.10 the inequality you are writting could be false if m < 2n (the exponent would be negative). You are also assuming x is an integer. this is clear if m>= 2n as x = n^{m/n - 2}, nut not if m
Can't use induction since x is not an integer. Have to use analysis with f(x)=2^x-4x.
Stunning math work! Only comment I feel urged to make is: love what you are doing here and wonder is (n,m) = (1,1) also a worthy paradigm shift?
Hey! Some years ago I stated a somewhat similar problem:
Find all triples m,n,k st mⁿ = (m+k)^(n-k), k≠0
The simplest is 2,4,2 but i recall finding an infinite set of solutions.
Is was 2018 and I offered a melon as a prize 😂
Edit: m,n,k are positive integers
Pretty clean
Wow I'm rusty, how are we getting the second implication where the difference of m and n is at least 0?
12:43
Michael Penn but every time he says "M" or "N" he gets faster...
Assuming 0 being excluded from N does (0,0) work if it had being allowed?
0^0=0^(0-0) iirc
If you're a "0^0=1" chad it works just fine apart from the arbitrary "solutions must be natural" restriction. If your a "0^0 is undefined" virgin then you can't have that solution.
0^0 is undefined
@@rukanee2883A lot of subfields (the majority of them, IIRC) define 0^0 to be 1, as it aligns well with lots of things, doesn't seem to cause any contradictions, and makes exponentiation total over the Natural Numbers including 0.
x is rational; not necessarily natural. It seems to me that all rational numbers between 2 and 5 are possible...
Can't x be 1 < x < 5? Need to show x is an integer.
Yes, it isn't shown at all. This means that the demonstration is not complete, or possibly invalid :-/ 😮😢
Outstanding sir! Very thanks for presentation of this nice maths
0:30 The problem being diophantine implies that n^(m-n) is a natural number. m and m^n being natural don't imply that.
m=1, n=2 results in n^(m-n)=1/2 not being natural. It is not true to say n, m are integers implies n^(m-n) is an integer.
There was already an initial condition that m, n are elements of natural number, so we do not have to complicate things.
@@rukanee2883 m & n being natural does not prevent m=n. Indeed, n^(m-n) being natural was used to justify m>=n. Thus, as stated, m and m^n being natural doesn't imply that n^(m-n) is natural.
I see in your country 0 is not natural ;)
I think he leaves it in sometimes, but just not today.
Is it a coincidence that - part from (1,1) - m is always n^2?
Part from (1,1) and (8,2), yes
Wdym part from? 1^2 = 1
Also not exactly a coincidence.
We’re looking at an equation where natural number variables satisfy an exponential equation. While it can’t be guaranteed by this reasoning, it shouldn’t be surprising to find that one solution is a natural exponent of the other.
Good on you for asking questions, though! That’s always important in mathematics.
Lowercase m,n is an unfortune choice of letters, especially with your handwriting, Michael. The sound is also similar. Perhaps it is just me, English is not my native language.
It’s just the standard convention where when you are working with Natural numbers your arbitrary variables are typically n and m. (And then if you are summing or multiplying over a range of Naturals then the variable used in the sum/product is k or j ranging from, say, 1 to n.).
So while m and n aren’t necessarily actually the best choice of letters in terms of clarity depending on handwriting and sounding similar when spoken, it’s just kind of tradition that people seem to use them. 🤷♂️
Take m=n=1
1¹ = 1⁰ = 1 😂
9:50 Zero anyway is not a solution, because plugging 0 at the original equation you either get 0=1, which is not true, or 0^0=0^0 which is also not true (since 0^0 cannot be evaluated)
Isn't m=n=0 also a valid solution? 0^0 = 0^(0-0) = 1.
Edit: For anyone replying to this with "0^0 is undefined/indefinite/etc.": It is only indefinite/undefined/whatever if you're talking about the limit of two functions, f(x)^g(x), where the limit as x->c is zero for both. Evaluated as a _number,_ 0^0 = 1. It has to. Otherwise, e^x is not equal to its Taylor series expansion at 0.
Seems natural to me, though it appears many definitions appear not to equate natural numbers with whole numbers.
0^0 is not agreed to equaling 1.
@@forcelifeforce That is the only value that it can consistently have, and in set theory and integer exponentiation--which is what matters in this context--it _is_ universally agreed to be 1. No other value matters.
@@TomFarrell-p9z I always forget that different groups define it differently. The way I was taught, 0 is a natural number. It seems Professor Penn falls in the camp that excludes 0.
@@ZekeRaidenyeah it’s often an American vs European split.
You miss a case for m=n. m=n=0 is also a solution if you consider that 0^0 = 1
0 is not considered a natural number
M and N have to be greater than 0. It’s a condition of the problem
For me, N = [0;inf[. N*=]0;inf[
Except that 0^0 is undefined.
@@MrWarlls"Natural numbers" generally means using the peano axioms, which clearly state that 1 is the smallest natural number.
How x can be a fraction, its definetely an integer. If its not m and n's powers can't be equal
Bro reply pls😢
1st comment 😊