4. Express the following decimal numbers in the form of [tex][tex]$\frac{p}{q}$[/tex][/tex] and write the prime factors of [tex]q[/tex]. What do you observe?

(i) 43.123
(ii) [tex][tex]$0.120112001120001 \ldots$[/tex][/tex]
(iii) [tex][tex]$43 . \overline{12}$[/tex][/tex]
(iv) [tex][tex]$0 . \overline{63}$[/tex][/tex]



Answer :

Let's go through each decimal number step-by-step to express them as fractions in the form of [tex]\(\frac{p}{q}\)[/tex] and find the prime factors of the denominators [tex]\(q\)[/tex].

### (i) 43.123
To convert 43.123 to a fraction, we can express it as [tex]\(\frac{43123}{1000}\)[/tex]:

[tex]\[ 43.123 = \frac{43123}{1000} \][/tex]

Next, we simplify this fraction by finding the Greatest Common Divisor (GCD) of 43123 and 1000.

- The prime factors of 1000 are [tex]\(2^3 \times 5^3\)[/tex].
- The prime factors of 43123 need to be determined for simplification.

After checking, 43123 is a prime number.

Therefore, the simplest form of the fraction is:

[tex]\[ 43.123 = \frac{43123}{1000} \][/tex]

So, we have:

[tex]\[ \frac{p}{q} = \frac{43123}{1000} \][/tex]
The prime factors of 1000 are [tex]\(2, 5\)[/tex].

### (ii) [tex]\(0.120112001120001 \ldots\)[/tex]
This number seems to follow a complex repeating pattern. To simplify the understanding, let's look at a partial pattern.

If it is not straightforward to express the repeating pattern, the decimal might be a non-standard repeating decimal that can be challenging to simplify directly without further tools.

### (iii) [tex]\(43 . \overline{12}\)[/tex]
This decimal has a repeating part:

[tex]\[ 43.\overline{12} = 43 + 0.\overline{12} \][/tex]

First, convert [tex]\(0.\overline{12}\)[/tex] to a fraction:
Let [tex]\( x = 0.\overline{12} \)[/tex].

Multiply by 100 to shift the decimal point:
[tex]\[ 100x = 12.\overline{12} \][/tex]

Now, subtract the original [tex]\(x\)[/tex] from this:
[tex]\[ 100x - x = 12.\overline{12} - 0.\overline{12} \][/tex]
[tex]\[ 99x = 12 \][/tex]
[tex]\[ x = \frac{12}{99} \][/tex]
[tex]\[ x = \frac{4}{33} \][/tex]

Thus,
[tex]\[ 43.\overline{12} = 43 + \frac{4}{33} \][/tex]
Convert the whole number part:
[tex]\[ = \frac{43 \cdot 33 + 4}{33} = \frac{1419 + 4}{33} = \frac{1423}{33} \][/tex]

So,
[tex]\[ \frac{p}{q} = \frac{1423}{33} \][/tex]
The prime factors of 33 are [tex]\(3, 11\)[/tex].

### (iv) [tex]\(0 . \overline{63}\)[/tex]
This decimal also has a repeating part:

[tex]\[ 0.\overline{63} \][/tex]

Let [tex]\( y = 0.\overline{63} \)[/tex].

Multiply by 100 to shift the decimal point:
[tex]\[ 100y = 63.\overline{63} \][/tex]

Now, subtract the original [tex]\(y\)[/tex] from this:
[tex]\[ 100y - y = 63.\overline{63} - 0.\overline{63} \][/tex]
[tex]\[ 99y = 63 \][/tex]
[tex]\[ y = \frac{63}{99} \][/tex]
[tex]\[ y = \frac{7}{11} \][/tex]

So,
[tex]\[ \frac{p}{q} = \frac{7}{11} \][/tex]
The prime factors of 11 are [tex]\(11\)[/tex] (as 11 is a prime number).

### Summary of Observations:
- (i) 43.123: [tex]\(\frac{43123}{1000}\)[/tex] with prime factors of the denominator being [tex]\(2\)[/tex] and [tex]\(5\)[/tex].
- (ii) [tex]\(0.120112001120001 \ldots\)[/tex]: (Pattern needs further clarification; prime factors calculation omitted due to complexity.)
- (iii) [tex]\(43.\overline{12}\)[/tex]: [tex]\(\frac{1423}{33}\)[/tex] with prime factors [tex]\(3\)[/tex] and [tex]\(11\)[/tex].
- (iv) [tex]\(0.\overline{63}\)[/tex]: [tex]\(\frac{7}{11}\)[/tex] with the prime factor [tex]\(11\)[/tex].

We observe that repeating decimals convert to fractions whose denominators often have simple prime factorizations. Non-repeating decimals usually have denominators that are powers of 10. For complex repeating patterns, the situation can be less straightforward and may need specialized methods for simplification.