Certainly! Let's break down the process of converting the fraction [tex]\(\frac{3}{7}\)[/tex] into a decimal.
1. Division: We'll divide 3 by 7 to find the equivalent decimal. Our task is then to determine how many times 7 fits into 3.
2. Initial Step: Since 7 doesn't go into 3 any whole number of times (7 is larger than 3), we start with 0. and proceed with decimal places.
3. First Decimal Place:
- We append a 0 to 3, turning it into 30.
- 7 fits into 30 four times (because [tex]\(7 \times 4 = 28\)[/tex]), leaving a remainder of 2 (since [tex]\(30 - 28 = 2\)[/tex]).
4. Second Decimal Place:
- Append another 0 to 2, turning it into 20.
- 7 fits into 20 two times (because [tex]\(7 \times 2 = 14\)[/tex]), leaving a remainder of 6 (since [tex]\(20 - 14 = 6\)[/tex]).
5. Third Decimal Place:
- Append a 0 to 6, turning it into 60.
- 7 fits into 60 eight times (because [tex]\(7 \times 8 = 56\)[/tex]), leaving a remainder of 4 (since [tex]\(60 - 56 = 4\)[/tex]).
6. Fourth Decimal Place:
- Append a 0 to 4, turning it into 40.
- 7 fits into 40 five times (because [tex]\(7 \times 5 = 35\)[/tex]), leaving a remainder of 5 (since [tex]\(40 - 35 = 5\)[/tex]).
7. Fifth Decimal Place:
- Append a 0 to 5, turning it into 50.
- 7 fits into 50 seven times (because [tex]\(7 \times 7 = 49\)[/tex]), leaving a remainder of 1 (since [tex]\(50 - 49 = 1\)[/tex]).
8. Sixth Decimal Place:
- Append a 0 to 1, turning it into 10.
- 7 fits into 10 one time (because [tex]\(7 \times 1 = 7\)[/tex]), leaving a remainder of 3 (since [tex]\(10 - 7 = 3\)[/tex]).
Notice here that we're back to where we started with a remainder of 3. This shows that the decimal pattern [tex]\(0.428571\)[/tex] will repeat.
Therefore, the decimal equivalent of [tex]\(\frac{3}{7}\)[/tex] is:
[tex]\[
0.42857142857142855 \ldots
\][/tex]
This fraction converts to a repeating decimal with the sequence "428571" repeating indefinitely.
