Prove $log_{4}6$ is irrational
$begingroup$
Thanks for taking the time to verify my approach and as well as my answer.
Background:
- B.S. in Business from a 4-year university taking CS courses online
- I would like some help with a basic proof from MIT's 6.042J Mathematics for Computer Science course.
The question is:
Prove $log_{4}6$ is irrational.
We prove the contradiction.
- Suppose $log_{4}6$ is rational (i.e. a quotient of integers)
$$log_{4}6 = m/n$$
- So we must have m, n integers without common prime factors such that
$$4^{m/n} = 6$$
- We will show that m and n are both even
$$(4^{m/n})^{n} = 6^{n}$$
- So
$$4^{m} = 6^{n}$$
- We then divide the two base numbers by their common factor, $2$, which gives us:
$$2^{m} = 3^{n}$$
- Since the product of two even numbers must be even AND the product of two odd numbers must be odd, $2^{m}$ and $3^{n}$ are not equivalent and therefore $m/n$ must not be rational.
Q.E.D. We conclude that $log_{4}6$ is irrational.
proof-verification logarithms irrational-numbers
edited Dec 13 '18 at 16:13
José Carlos Santos
173k23133241
asked Dec 13 '18 at 16:07
BryceBryce
141
$endgroup$
add a comment |
$begingroup$
Thanks for taking the time to verify my approach and as well as my answer.
Background:
- B.S. in Business from a 4-year university taking CS courses online
- I would like some help with a basic proof from MIT's 6.042J Mathematics for Computer Science course.
The question is:
Prove $log_{4}6$ is irrational.
We prove the contradiction.
- Suppose $log_{4}6$ is rational (i.e. a quotient of integers)
$$log_{4}6 = m/n$$
- So we must have m, n integers without common prime factors such that
$$4^{m/n} = 6$$
- We will show that m and n are both even
$$(4^{m/n})^{n} = 6^{n}$$
- So
$$4^{m} = 6^{n}$$
- We then divide the two base numbers by their common factor, $2$, which gives us:
$$2^{m} = 3^{n}$$
- Since the product of two even numbers must be even AND the product of two odd numbers must be odd, $2^{m}$ and $3^{n}$ are not equivalent and therefore $m/n$ must not be rational.
Q.E.D. We conclude that $log_{4}6$ is irrational.
proof-verification logarithms irrational-numbers
edited Dec 13 '18 at 16:13
José Carlos Santos
173k23133241
asked Dec 13 '18 at 16:07
BryceBryce
141
$endgroup$
1
$begingroup$
Until $4^m=6^n$, you are ok, but after that you go astray. To continue on the path you started on, from $4^m=6^n$, deduce $2^{2m}=2^n3^n$ hence $2^{2m-n}=3^n$. And now, how would you conclude?
$endgroup$
– Did
Dec 13 '18 at 16:13
add a comment |
$begingroup$
Thanks for taking the time to verify my approach and as well as my answer.
Background:
- B.S. in Business from a 4-year university taking CS courses online
- I would like some help with a basic proof from MIT's 6.042J Mathematics for Computer Science course.
The question is:
Prove $log_{4}6$ is irrational.
We prove the contradiction.
- Suppose $log_{4}6$ is rational (i.e. a quotient of integers)
$$log_{4}6 = m/n$$
- So we must have m, n integers without common prime factors such that
$$4^{m/n} = 6$$
- We will show that m and n are both even
$$(4^{m/n})^{n} = 6^{n}$$
- So
$$4^{m} = 6^{n}$$
- We then divide the two base numbers by their common factor, $2$, which gives us:
$$2^{m} = 3^{n}$$
- Since the product of two even numbers must be even AND the product of two odd numbers must be odd, $2^{m}$ and $3^{n}$ are not equivalent and therefore $m/n$ must not be rational.
Q.E.D. We conclude that $log_{4}6$ is irrational.
proof-verification logarithms irrational-numbers
edited Dec 13 '18 at 16:13
José Carlos Santos
173k23133241
asked Dec 13 '18 at 16:07
BryceBryce
141
$endgroup$
Thanks for taking the time to verify my approach and as well as my answer.
Background:
- B.S. in Business from a 4-year university taking CS courses online
- I would like some help with a basic proof from MIT's 6.042J Mathematics for Computer Science course.
The question is:
Prove $log_{4}6$ is irrational.
We prove the contradiction.
- Suppose $log_{4}6$ is rational (i.e. a quotient of integers)
$$log_{4}6 = m/n$$
- So we must have m, n integers without common prime factors such that
$$4^{m/n} = 6$$
- We will show that m and n are both even
$$(4^{m/n})^{n} = 6^{n}$$
- So
$$4^{m} = 6^{n}$$
- We then divide the two base numbers by their common factor, $2$, which gives us:
$$2^{m} = 3^{n}$$
- Since the product of two even numbers must be even AND the product of two odd numbers must be odd, $2^{m}$ and $3^{n}$ are not equivalent and therefore $m/n$ must not be rational.
Q.E.D. We conclude that $log_{4}6$ is irrational.
proof-verification logarithms irrational-numbers
proof-verification logarithms irrational-numbers
edited Dec 13 '18 at 16:13
José Carlos Santos
173k23133241
asked Dec 13 '18 at 16:07
BryceBryce
141
edited Dec 13 '18 at 16:13
José Carlos Santos
173k23133241
asked Dec 13 '18 at 16:07
BryceBryce
141
edited Dec 13 '18 at 16:13
José Carlos Santos
173k23133241
edited Dec 13 '18 at 16:13
José Carlos Santos
173k23133241
edited Dec 13 '18 at 16:13
José Carlos Santos
173k23133241
173k23133241
asked Dec 13 '18 at 16:07
BryceBryce
141
asked Dec 13 '18 at 16:07
BryceBryce
141
asked Dec 13 '18 at 16:07
BryceBryce
141
141
1
$begingroup$
Until $4^m=6^n$, you are ok, but after that you go astray. To continue on the path you started on, from $4^m=6^n$, deduce $2^{2m}=2^n3^n$ hence $2^{2m-n}=3^n$. And now, how would you conclude?
$endgroup$
– Did
Dec 13 '18 at 16:13
add a comment |
1
$begingroup$
Until $4^m=6^n$, you are ok, but after that you go astray. To continue on the path you started on, from $4^m=6^n$, deduce $2^{2m}=2^n3^n$ hence $2^{2m-n}=3^n$. And now, how would you conclude?
$endgroup$
– Did
Dec 13 '18 at 16:13
1
1
$begingroup$
Until $4^m=6^n$, you are ok, but after that you go astray. To continue on the path you started on, from $4^m=6^n$, deduce $2^{2m}=2^n3^n$ hence $2^{2m-n}=3^n$. And now, how would you conclude?
$endgroup$
– Did
Dec 13 '18 at 16:13
$begingroup$
Until $4^m=6^n$, you are ok, but after that you go astray. To continue on the path you started on, from $4^m=6^n$, deduce $2^{2m}=2^n3^n$ hence $2^{2m-n}=3^n$. And now, how would you conclude?
$endgroup$
– Did
Dec 13 '18 at 16:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It's fine until you reach that equality $4^m=6^n$. But you can't just divide by their common factor $2$. Are you dividing by $2^m$ or by $2^n$?
You can say that, since $n>0$, then $3mid6^n$. But $3nmid4^m$.
answered Dec 13 '18 at 16:12
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
add a comment |
$begingroup$
Your proof works until you reach $4^m=6^n.$ However, you don't really need to divide by any factor of $2.$
Instead, you can add:
Since $log_46$ is positive, as 6 is greater than 4, $m$ and $n$ are positive integers. This means that the right hand side ($6^n$) is divisible by $3.$ However,
the left hand side, $4^m$ can be prime factorized as $2^{2m},$ and therefore is not divisible by $3,$ meaning that $4^mneq 6^n,$ yielding contradiction.
answered Dec 13 '18 at 17:26
BetaGammaBetaGamma
211
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
It's fine until you reach that equality $4^m=6^n$. But you can't just divide by their common factor $2$. Are you dividing by $2^m$ or by $2^n$?
You can say that, since $n>0$, then $3mid6^n$. But $3nmid4^m$.
answered Dec 13 '18 at 16:12
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
add a comment |
$begingroup$
It's fine until you reach that equality $4^m=6^n$. But you can't just divide by their common factor $2$. Are you dividing by $2^m$ or by $2^n$?
You can say that, since $n>0$, then $3mid6^n$. But $3nmid4^m$.
answered Dec 13 '18 at 16:12
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
add a comment |
$begingroup$
It's fine until you reach that equality $4^m=6^n$. But you can't just divide by their common factor $2$. Are you dividing by $2^m$ or by $2^n$?
You can say that, since $n>0$, then $3mid6^n$. But $3nmid4^m$.
answered Dec 13 '18 at 16:12
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
It's fine until you reach that equality $4^m=6^n$. But you can't just divide by their common factor $2$. Are you dividing by $2^m$ or by $2^n$?
You can say that, since $n>0$, then $3mid6^n$. But $3nmid4^m$.
answered Dec 13 '18 at 16:12
José Carlos SantosJosé Carlos Santos
173k23133241
answered Dec 13 '18 at 16:12
José Carlos SantosJosé Carlos Santos
173k23133241
answered Dec 13 '18 at 16:12
José Carlos SantosJosé Carlos Santos
173k23133241
answered Dec 13 '18 at 16:12
José Carlos SantosJosé Carlos Santos
173k23133241
173k23133241
add a comment |
add a comment |
$begingroup$
Your proof works until you reach $4^m=6^n.$ However, you don't really need to divide by any factor of $2.$
Instead, you can add:
Since $log_46$ is positive, as 6 is greater than 4, $m$ and $n$ are positive integers. This means that the right hand side ($6^n$) is divisible by $3.$ However,
the left hand side, $4^m$ can be prime factorized as $2^{2m},$ and therefore is not divisible by $3,$ meaning that $4^mneq 6^n,$ yielding contradiction.
answered Dec 13 '18 at 17:26
BetaGammaBetaGamma
211
$endgroup$
add a comment |
$begingroup$
Your proof works until you reach $4^m=6^n.$ However, you don't really need to divide by any factor of $2.$
Instead, you can add:
Since $log_46$ is positive, as 6 is greater than 4, $m$ and $n$ are positive integers. This means that the right hand side ($6^n$) is divisible by $3.$ However,
the left hand side, $4^m$ can be prime factorized as $2^{2m},$ and therefore is not divisible by $3,$ meaning that $4^mneq 6^n,$ yielding contradiction.
answered Dec 13 '18 at 17:26
BetaGammaBetaGamma
211
$endgroup$
add a comment |
$begingroup$
Your proof works until you reach $4^m=6^n.$ However, you don't really need to divide by any factor of $2.$
Instead, you can add:
Since $log_46$ is positive, as 6 is greater than 4, $m$ and $n$ are positive integers. This means that the right hand side ($6^n$) is divisible by $3.$ However,
the left hand side, $4^m$ can be prime factorized as $2^{2m},$ and therefore is not divisible by $3,$ meaning that $4^mneq 6^n,$ yielding contradiction.
answered Dec 13 '18 at 17:26
BetaGammaBetaGamma
211
$endgroup$
Your proof works until you reach $4^m=6^n.$ However, you don't really need to divide by any factor of $2.$
Instead, you can add:
Since $log_46$ is positive, as 6 is greater than 4, $m$ and $n$ are positive integers. This means that the right hand side ($6^n$) is divisible by $3.$ However,
the left hand side, $4^m$ can be prime factorized as $2^{2m},$ and therefore is not divisible by $3,$ meaning that $4^mneq 6^n,$ yielding contradiction.
answered Dec 13 '18 at 17:26
BetaGammaBetaGamma
211
answered Dec 13 '18 at 17:26
BetaGammaBetaGamma
211
answered Dec 13 '18 at 17:26
BetaGammaBetaGamma
211
answered Dec 13 '18 at 17:26
BetaGammaBetaGamma
211
211
add a comment |
add a comment |
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$begingroup$
Until $4^m=6^n$, you are ok, but after that you go astray. To continue on the path you started on, from $4^m=6^n$, deduce $2^{2m}=2^n3^n$ hence $2^{2m-n}=3^n$. And now, how would you conclude?
$endgroup$
– Did
Dec 13 '18 at 16:13