Proof that $log_q(p)$ is (mostly) irrational for integers q,p
$begingroup$
Does this proof make sense, especially the last part?
Take $q>p$. We consider the equation $p^aq^b=p^{a'} q^{b'}$. Given that $a,b$ and $a',b'$ are distinct respectively. We solve, realising that $frac{a-a'}{b'-b} = log_p(q)$. Notice that the existence of solutions implies that the latter is rational. Also, in the same way, a solution for the second equation implies a solution for the first. Therefore solutions to
$p^aq^b=p^{a'} q^{b'}$ is equivalent to saying that $log_p(q)$ is rational.
If $p,q$ don't share the same prime factors, it is easy to see that there exist no distinct solutions.
Now consider the case in which they do. The multiplicity of a prime k on both sides is $k_p a + k_q b = k_p a' + k_q b'$ so $$frac{k_q}{k_p} = log_p(q)$$
This must be a constant and it must be the same for any prime factor $k$. Therefore $q = p^r$ where $r$ is a rational number. Or solutions exist $q^b = p^a$ . Can we simplify this condition further?
Alternatively we could assume $log_p(q) = frac{a}{b}$ is injective so $log_p(p^frac{a}{b})=log_p(q)$ so $p^a = q^b$. This leads us to the same condition.
I'm not sure about the clarity or validity of proof 1. Proof 2 seems cleaner but it assumes injectivity. Could someone help me clear this up so it is concise and conceptually clear?
proof-verification proof-writing logarithms irrational-numbers
asked Dec 30 '18 at 9:37
Dis-integratingDis-integrating
1,043526
$endgroup$
add a comment |
$begingroup$
Does this proof make sense, especially the last part?
Take $q>p$. We consider the equation $p^aq^b=p^{a'} q^{b'}$. Given that $a,b$ and $a',b'$ are distinct respectively. We solve, realising that $frac{a-a'}{b'-b} = log_p(q)$. Notice that the existence of solutions implies that the latter is rational. Also, in the same way, a solution for the second equation implies a solution for the first. Therefore solutions to
$p^aq^b=p^{a'} q^{b'}$ is equivalent to saying that $log_p(q)$ is rational.
If $p,q$ don't share the same prime factors, it is easy to see that there exist no distinct solutions.
Now consider the case in which they do. The multiplicity of a prime k on both sides is $k_p a + k_q b = k_p a' + k_q b'$ so $$frac{k_q}{k_p} = log_p(q)$$
This must be a constant and it must be the same for any prime factor $k$. Therefore $q = p^r$ where $r$ is a rational number. Or solutions exist $q^b = p^a$ . Can we simplify this condition further?
Alternatively we could assume $log_p(q) = frac{a}{b}$ is injective so $log_p(p^frac{a}{b})=log_p(q)$ so $p^a = q^b$. This leads us to the same condition.
I'm not sure about the clarity or validity of proof 1. Proof 2 seems cleaner but it assumes injectivity. Could someone help me clear this up so it is concise and conceptually clear?
proof-verification proof-writing logarithms irrational-numbers
asked Dec 30 '18 at 9:37
Dis-integratingDis-integrating
1,043526
$endgroup$
add a comment |
$begingroup$
Does this proof make sense, especially the last part?
Take $q>p$. We consider the equation $p^aq^b=p^{a'} q^{b'}$. Given that $a,b$ and $a',b'$ are distinct respectively. We solve, realising that $frac{a-a'}{b'-b} = log_p(q)$. Notice that the existence of solutions implies that the latter is rational. Also, in the same way, a solution for the second equation implies a solution for the first. Therefore solutions to
$p^aq^b=p^{a'} q^{b'}$ is equivalent to saying that $log_p(q)$ is rational.
If $p,q$ don't share the same prime factors, it is easy to see that there exist no distinct solutions.
Now consider the case in which they do. The multiplicity of a prime k on both sides is $k_p a + k_q b = k_p a' + k_q b'$ so $$frac{k_q}{k_p} = log_p(q)$$
This must be a constant and it must be the same for any prime factor $k$. Therefore $q = p^r$ where $r$ is a rational number. Or solutions exist $q^b = p^a$ . Can we simplify this condition further?
Alternatively we could assume $log_p(q) = frac{a}{b}$ is injective so $log_p(p^frac{a}{b})=log_p(q)$ so $p^a = q^b$. This leads us to the same condition.
I'm not sure about the clarity or validity of proof 1. Proof 2 seems cleaner but it assumes injectivity. Could someone help me clear this up so it is concise and conceptually clear?
proof-verification proof-writing logarithms irrational-numbers
asked Dec 30 '18 at 9:37
Dis-integratingDis-integrating
1,043526
$endgroup$
Does this proof make sense, especially the last part?
Take $q>p$. We consider the equation $p^aq^b=p^{a'} q^{b'}$. Given that $a,b$ and $a',b'$ are distinct respectively. We solve, realising that $frac{a-a'}{b'-b} = log_p(q)$. Notice that the existence of solutions implies that the latter is rational. Also, in the same way, a solution for the second equation implies a solution for the first. Therefore solutions to
$p^aq^b=p^{a'} q^{b'}$ is equivalent to saying that $log_p(q)$ is rational.
If $p,q$ don't share the same prime factors, it is easy to see that there exist no distinct solutions.
Now consider the case in which they do. The multiplicity of a prime k on both sides is $k_p a + k_q b = k_p a' + k_q b'$ so $$frac{k_q}{k_p} = log_p(q)$$
This must be a constant and it must be the same for any prime factor $k$. Therefore $q = p^r$ where $r$ is a rational number. Or solutions exist $q^b = p^a$ . Can we simplify this condition further?
Alternatively we could assume $log_p(q) = frac{a}{b}$ is injective so $log_p(p^frac{a}{b})=log_p(q)$ so $p^a = q^b$. This leads us to the same condition.
I'm not sure about the clarity or validity of proof 1. Proof 2 seems cleaner but it assumes injectivity. Could someone help me clear this up so it is concise and conceptually clear?
proof-verification proof-writing logarithms irrational-numbers
proof-verification proof-writing logarithms irrational-numbers
asked Dec 30 '18 at 9:37
Dis-integratingDis-integrating
1,043526
asked Dec 30 '18 at 9:37
Dis-integratingDis-integrating
1,043526
edited Dec 30 '18 at 9:43
Dis-integrating
asked Dec 30 '18 at 9:37
Dis-integratingDis-integrating
1,043526
asked Dec 30 '18 at 9:37
Dis-integratingDis-integrating
1,043526
asked Dec 30 '18 at 9:37
Dis-integratingDis-integrating
1,043526
1,043526
add a comment |
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