How can $\frac{4}{3} \times 3=4$ if $ \frac{4}{3}$ is $1.3$?

Question

Ok use your closest calculator, and type $\frac{4}{3}$, which is $1.3333333333$,and then multiply it with $3$ which is $3.9999999999$ but then type $\frac{4}{3} \times 3=4$ how?. How can it be $4$ if $\frac{4}{3}$ is $1.3333333333$ and when you multiply it with $3$ is $3.9999999999$.

Answer 1

The fallacy is that you only consider finite many $3's$ giving only finite many $9's$ , when multiplied by $3$. But we have infinite many $3's$ , and the result is exactly $4$. The reason is exactly the same why $$0.999\cdots =1$$ which you can show with a geometric series with start value $\frac{9}{10}$ and quotient $\frac{1}{10}$. Using the formula, you get exactly $1$.

Answer 2

Rational numbers are numbers that can be expressed as $\frac{p}{q}$, where p and q are relatively prime integers. Notice that the definition does not mention the decimal expansion. The decimal representation of $\frac{4}{3}$ is defined as $1+\sum_{i=1}^{\infty}\frac{3}{10^i}$. Multiplying by 3, the decimal expansion for $\frac{4}{3}*3$ becomes $3+\sum_{i=1}^\infty\frac{9}{10^i}$. The value of an infinite sum is defined as the limit of the partial sums, or, in simpler terms, what number is approached when we sum up the first $n$ terms, followed by the first $n+1$ term, etc. Doing that here, we see the sequence is $.9$, $.99$, $.999$, etc. It should be obvious that we are getting closer and closer to $1$. Thus, $\sum_{i=1}^{\infty}\frac{3}{10^i}=1$. We now have $\frac{4}{3}*3=1+\sum_{i=1}^{\infty}\frac{3}{10^i}=1+3=4$.

Answer 3

The calculators usually keep invisible extra decimals (so-called guard digits) and round for display.

For example, computing on $10$ digits,

$$\frac43=1.333333333$$ then

$$\frac43\, 3=3.999999999$$

Now, displayed on $8$ digits with rounding,

$$4.0000000$$ or simply $$4$$

On some calculators,

$$4\div3\times3-4$$ might not return $0$, even if the intermediate result $4$ is shown.

Answer 4

Please note that $$\frac4 3 \neq 1.\underbrace{333333333...3}_{n}, $$ no matter how large $n$ is.