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To say that $\log_2 3$ is rational would mean that there are integers $m$, $n$ such that $\log_2 3=\dfrac mn$ with $n \ne 0$. That would imply that $2^{m/n}=3$, so that $2^m=3^n$. But that says an even number equals an odd number, which is impossible, so $\log_2 3$ cannot be rational.
- Do logarithms ever produce rational numbers?
Logb (n) where n is a power of b produces a rational number;...
- Prove that $\\log_2 3$ is irrational - Mathematics Stack Exchange
Let log2 3 = p/q log 2 3 = p / q where p ∈ Z p ∈ Z and q ∈N...
- Do logarithms ever produce rational numbers?
14 Μαρ 2023 · Logb (n) where n is a power of b produces a rational number; for example; Log2 (8) = 3. But, Log2 (3) = 1.5849625007211561814537389439478165087598144076924810604557526545... Thus, Log2 (3) produces an irrational number.
Introduction to Logarithms. In its simplest form, a logarithm answers the question: How many of one number multiply together to make another number? Example: How many 2 s multiply together to make 8? Answer: 2 × 2 × 2 = 8, so we had to multiply 3 of the 2 s to get 8. So the logarithm is 3. How to Write it. We write it like this: log2(8) = 3.
Let log2 3 = p/q log 2 3 = p / q where p ∈ Z p ∈ Z and q ∈N q ∈ N (since surely log2 3> 0 log 2 3> 0 you may directly assume that p ∈N p ∈ N as well.) Now it must hold. 2p = 3q 2 p = 3 q. But note that one side is even and the other one is odd! Hence log2 3 log 2 3 is not rational!
Assume that \( \log_{2} 3 \) is a rational number. Therefore, it can be written as a ratio of two integers, say \( \log_{2} 3 = \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q \e 0 \).
Proof. Observe that if pr divides a number in {1,2,...,n}, then pr ≤ n so that r ≤ logn/logp. On the other hand, p[logn/logp] does divide one such number (namely itself). Thus, d n = Y p≤n p[logn/logp]. Explain how the rest follows from the Prime Number Theorem. Theorem 9. The number log2 is irrational. Proof. Assume log2 = a/b for some ...
A number is rational if it is in the form \dfrac {p} {q} qp, where p,q p,q are integers (q \ne 0 q = 0). Piecing this together, we want to show that \log_2 3 log23 is irrational; i.e. that it can't be written in the form \dfrac {p} {q} qp for any integers p,q p,q.
