a student claims: \repeating decimals can be turned into fractions the same way as terminating decimals.\ do…

Explanation:

Step1: Convert Terminating Decimal (0.75)

To convert ( 0.75 ) to a fraction, we use the place value. Since ( 0.75 ) is ( 75 ) hundredths, we can write it as ( \frac{75}{100} ), and simplify to ( \frac{3}{4} ). Here, multiplying by ( 10^2 = 100 ) (because there are 2 decimal places) allows us to move the decimal right 2 places to get an integer numerator.

Step2: Attempt Terminating Method on Repeating (0.333...)

If we try the terminating decimal method on ( 0.333... ), we see that the number of decimal places is infinite (it repeats forever), so we can't multiply by a finite power of 10 to eliminate the decimal. For example, ( 0.333... \times 10 = 3.333... ), ( 0.333... \times 100 = 33.333... ), and so on—there's always a repeating decimal part left.

Step3: Convert Repeating Decimal (0.333...)

Let ( x = 0.333... ). Multiply both sides by 10: ( 10x = 3.333... ). Now subtract the original equation (( x = 0.333... )) from this new equation: ( 10x - x = 3.333... - 0.333... ). This simplifies to ( 9x = 3 ). Then, solve for ( x ): ( x = \frac{3}{9} = \frac{1}{3} ). The subtraction eliminates the repeating decimal part, allowing us to solve for ( x ) (the fraction equivalent).

Answer:

I disagree. For the terminating decimal ( 0.75 ), we use a finite power of 10 (e.g., ( 10^2 = 100 )): ( 0.75=\frac{75}{100}=\frac{3}{4} ), as the decimal has a finite number of places. For the repeating decimal ( 0.333... ), a finite power of 10 can’t eliminate the repeating part (e.g., ( 10 \times 0.333... = 3.333... ), still repeating). To convert ( 0.333... ), let ( x = 0.333... ), so ( 10x = 3.333... ). Subtracting ( x ) from ( 10x ): ( 10x - x = 3.333... - 0.333... ), which gives ( 9x = 3 ), so ( x=\frac{1}{3} ). Thus, repeating decimals require a different method (involving subtraction to eliminate the repeating part) than terminating decimals (using a finite power of 10).