Question

Докажите ирроциональность числа log3 4

Answer 1

㏒₃4 > 0, т.к. основание равно 3 > 1, а подлогарифмическое выражение  равно 4, то есть его значение больше значения основания.

Допустим, что это число рационально. Значит оно представимо в виде b/n, где b/n > 0,  b, n - целые, b, n ≠ 0.  Не нарушая общности, допустим, что b, n - натуральные.

Тогда:

㏒₃4=b/n → n*㏒₃4=b → ㏒₃(4ⁿ)=b → 3ᵇ=4ⁿ

3ᵇ - нечетное для любой натуральной степени b [3ᵇ≡1ᵇ(mod 2)=1]

4ⁿ - четное для любой натуральной степени b [4ᵇ≡0ᵇ(mod 2)=0]

Получаем равенство четного и нечетного чисел. Противоречие. Значит число ㏒₃4 иррационально.

Ч.т.д.

Answer 2

To prove the irrationality of the number $\log_3 4$, we can use a proof by contradiction.

Assume that $\log_3 4$ is rational, which means it can be expressed as a fraction $\frac{a}{b}$ where $a$ and $b$ are integers with no common factors other than 1, and $b$ is not equal to 0.

So, we have: $\log_3 4 = \frac{a}{b}$

Now, let's rewrite the logarithmic equation in exponential form. For any positive numbers $x$ and $y$ where $x > 1$, if $\log_x y = \frac{a}{b}$, then $x^\frac{a}{b} = y$.

Applying this to our equation, we have: $3^\frac{a}{b} = 4$

Now, raise both sides of the equation to the power of $b$: $3^a = 4^b$

At this point, we have a contradiction. On the left side, we have $3^a$, which is a multiple of 3, since $3$ is a factor of the base $3$. On the right side, we have $4^b$, which is always a power of 2 and does not have any factors of 3.

Since 3 is not a factor of 4, there is no way for both sides of the equation to be equal. This contradiction shows that our initial assumption that $\log_3 4$ is rational must be false.

Hence, we can conclude that $\log_3 4$ is an irrational number.