How to represent $(43.3)_7$ in base-8?

Question

I am trying to represent $(43.3)_7$ in base-8.

But in only one integer digit by truncating the rest and using the numerical unsigned representation.

Answer 1

Let's look at the definition of these things.

Interpreting

$$43.3_7 = 4\cdot 7^1 + 3\cdot 7^0 + 3\cdot 7^{-1} \approx 31.4286_{10}.$$

Now we change to base $8$. We have $8^2 = 64$ is too large, so the first digit we look at will be $8^1$. We have $31.4286/8 \approx 3.9285$ so the first digit will be a $3$.

Assuming by "one integer digit" you mean to truncate the result here, then the answer would be $3_8$.

Answer 2

As best I can read the second paragraph, you want one character (digit?) in the fractional part(beyond the decimal point?). $43_7=31_{10}=37_8$ for the integer part. For the fraction $\frac 37$ is closer to $\frac 38$ than $\frac 48$, so it would be $37.3_8$ is the closest.

Answer 3

I’m going to ignore the second paragraph, since it looks as if something has been omitted from it. The number in question, in elementary-school notation, is $31\frac{3}{7}$. Since the base-$8$ expansion of $1/7$ is $.1111\cdots$, in base-$8$, the number in question is $37.3333\cdots$. We leave it to OP to round.