Fractions & Equivalent Fractions

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Explain this!

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Why does multiplying a whole number by a fraction produce an equivalent fraction?

Example: 2/3 * 6 = 4 Divide both sides by 6, and this becomes 2/3 = 4/6.

Another example: 6/16 * 8 = 3 Divide both sides by 8, and this becomes 6/16 = 3/8.
 
Explain this! said:
Why does multiplying a whole number by a fraction produce an equivalent fraction?
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It does NOT. Multiplying the numerator and denominator by the same number does produce an equivalent fraction.
But your examples show something entirely different - multiplying both sides of an equality by the same number produces an equality.
 
Explain this! said:
Why does multiplying a whole number by a fraction produce an equivalent fraction?
Click to expand...
It doesn't! That sounds like you are saying that 2/3 * 6 = 4 is equivalent to 2/3.

Explain this! said:
Example: 2/3 * 6 = 4 Divide both sides by 6, and this becomes 2/3 = 4/6.

Another example: 6/16 * 8 = 3 Divide both sides by 8, and this becomes 6/16 = 3/8.
Click to expand...
What you're doing is multiplying and dividing 2/3 by 6, which of course leaves the value unchanged.
 
This is very strange. I would think that you would have learned early in your mathematics lessons that doing the same thing to both sides of a true equation, adding or subtracting the same number, multiplying or dividing both sides by the same thing, taking the square root of both sides or squaring both sides, etc., results in a true equation.

This because those operation are "well defined"- a+ b, for the same a and b, will give the same result no matter where or when you add them.

That is why we specifically define \(\displaystyle \sqrt{a}\) to be the positive number, x, such that \(\displaystyle x^2= a\). If we did not do that we might wind up saying, "since 4= 4, \(\displaystyle \sqrt{4}= \sqrt{4}\), \(\displaystyle 2= -2\)" which is not true. But with that definition, ""since 4= 4, \(\displaystyle \sqrt{4}= \sqrt{4}\), \(\displaystyle 2= 2\) which is true.

We like operations that are "well defined"!
 
Explain this! said:
Why does multiplying a whole number by a fraction produce an equivalent fraction?

Example: 2/3 * 6 = 4 Divide both sides by 6, and this becomes 2/3 = 4/6.

Another example: 6/16 * 8 = 3 Divide both sides by 8, and this becomes 6/16 = 3/8.
Click to expand...

You possibly mean -
"Why does multiplying a fraction by a whole number, 'cancelling' and then putting it over the whole number again, produce a 'different' but equivalent fraction?
(Here I am considering [MATH]\frac{2}{3}[/MATH] and [MATH]\frac{4}{6}[/MATH] 'different' fractions, in that they use different pairs of whole numbers, but they both represent the same real number, so are equivalent fractions).

Take a fraction [MATH]\frac{a}{b}[/MATH] and multiply it by a whole number c. You get [MATH]\frac{ac}{b}[/MATH][MATH]\begin{align*}\frac{a}{b} \times c &= \frac{ac}{b}\\ \div c \hspace2ex \frac{a}{b}&=\frac{\frac{ac}{b}}{c} \end {align*}[/MATH]Clearly this is only the 'same fraction' if [MATH]b=c[/MATH] and [MATH]\frac{ac}{b}[/MATH] is a whole number
Also, these must be equivalent fractions, since as pointed out by other posts, if you multiply by c and then divide by (non-zero) c, you get back to the number you started with.
 
Perhaps I did not word my question properly.

Find the missing number in 2/3 = ?/6. If 6 is multiplied by 2/3 the result is 4. 2/3 and 4/6 are equivalent. Why doess this calculation, 2/3 * 6, produce the value for?
 
lex said:
You possibly mean -
"Why does multiplying a fraction by a whole number, 'cancelling' and then putting it over the whole number again, produce a 'different' but equivalent fraction?
(Here I am considering [MATH]\frac{2}{3}[/MATH] and [MATH]\frac{4}{6}[/MATH] 'different' fractions, in that they use different pairs of whole numbers, but they both represent the same real number, so are equivalent fractions).

Take a fraction [MATH]\frac{a}{b}[/MATH] and multiply it by a whole number c. You get [MATH]\frac{ac}{b}[/MATH][MATH]\begin{align*}\frac{a}{b} \times c &= \frac{ac}{b}\\ \div c \hspace2ex \frac{a}{b}&=\frac{\frac{ac}{b}}{c} \end {align*}[/MATH]Clearly this is only the 'same fraction' if [MATH]b=c[/MATH] and [MATH]\frac{ac}{b}[/MATH] is a whole number
Also, these must be equivalent fractions, since as pointed out by other posts, if you multiply by c and then divide by (non-zero) c, you get back to the number you started with.
Click to expand...

Yes!
 
Explain this! said:
Find the missing number in 2/3 = ?/6. If 6 is multiplied by 2/3 the result is 4. 2/3 and 4/6 are equivalent. Why doess this calculation, 2/3 * 6, produce the value for?
Click to expand...
In that case it is just solving an equation to find [MATH]x[/MATH]:

[MATH]\frac{x}{6}=\frac{2}{3}[/MATH]
[MATH]\times 6[/MATH] on both sides:
[MATH]x=6\times \frac{2}{3}\hspace3ex (=4)[/MATH]
 
lex said:
In that case it is just solving an equation to find [MATH]x[/MATH]:

[MATH]\frac{x}{6}=\frac{2}{3}[/MATH]
[MATH]\times 6[/MATH] on both sides:
[MATH]x=6\times \frac{2}{3}\hspace3ex (=4)[/MATH]
Click to expand...
You just reversed what I am trying to understand. I can solve for a simple proportion as you indicated, but why does 2/3 * 6 produce that amount or number (4) that makes it equivalent to 2/3 as in 2/3 = 4/6?

It is more of a philological question than a mathematical one.
 
Explain this! said:
You just reversed what I am trying to understand. I can solve for a simple proportion as you indicated, but why does 2/3 * 6 produce that amount or number (4) that makes it equivalent to 2/3 as in 2/3 = 4/6?

It is more of a philological question than a mathematical one.
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Start with any 2 equivalent fractions, e.g.
2/3 = 4/6

Multiply both sides by one of the denominators. What do you get?
 
[MATH]\frac{2}{3} \times 6 = \boldsymbol{4}[/MATH]
Divide both sides by 6. I'm going to write the division by 6 slightly differently on each side:
[MATH]\frac{2}{3} \times 6 \div 6 = \frac{\boldsymbol{4}}{6}[/MATH]
[MATH]\times 6[/MATH] and [MATH]\div 6[/MATH] are inverse operations and have no overall effect on a number, so [MATH]\frac{2}{3} \times 6 \div 6[/MATH] is just [MATH]\frac{2}{3}[/MATH]Therefore:
[MATH]\frac{2}{3}=\frac{\boldsymbol{4}}{6}[/MATH](the 4 having come from [MATH]\frac{2}{3} \times 6 = \boldsymbol{4}[/MATH])

(In other words you have multiplied by 6 and divided by 6, so you have the same numerical answer [MATH]\frac{2}{3}[/MATH]and it is the same as [MATH]\frac{2}{3} \times 6[/MATH] on the top line of a fraction and 6 on the bottom line of the fraction).
 
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I see it this way: 2/3 * 6 = 2 * (1/3 * 6). 6 divided by 3 indicates how many times 3 divides into 6. It is 2 times. Now multiply 2 times 2 from the separated 2/3 (2 * 1/3) to get 4. For some reason, the two fractions remain proportional by this calculation 2/3 * 6 = 4. 2/3 = 4/6 are proportional.
 
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I still seem to be missing the mark here.
[MATH]\tfrac{2}{3} = \frac{\;\frac{2}{3}\;}{1}[/MATH]Multiply top and bottom by 6 to get an equivalent fraction:
[MATH]\frac{2}{3} = \frac{\frac{2}{3}\times 6}{1\times 6}=\frac{4}{6}[/MATH]
 
Explain this! said:
I see it this way: 2/3 * 6 = 2 * (1/3 * 6). 6 divided by 3 indicates how many times 3 divides into 6. It is 2 times. Now multiply 2 times 2 from the separated 2/3 (2 * 1/3) to get 4. For some reason, the two fractions remain proportional by this calculation 2/3 * 6 = 4. 2/3 = 4/6 are proportional.
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You are confusing yourself by mixing together instructions in English with operations in math.

You seem to have been taught the following procedure.

To find the equivalent fraction to 2/3 with a denominator of 24:
(1) Divide 24 by 3, getting 8;
(2) Multiply 2 by 8, getting 16;
(3) Put 16 over 24 and get the following: [MATH]\dfrac{16}{24} = \dfrac{2}{3}.[/MATH]
You want to know why it works from first principles.

[MATH]a \ne 0 \implies \dfrac{a}{a} = 1.[/MATH]
[MATH]b = b * 1.[/MATH]
Two basic principles of arithmetic.

What the English instructions do is to summarize the following sequence of arithmetic operations.

[MATH]\dfrac{2}{3} = \dfrac{2}{3} * 1 = \dfrac{2}{3} * \dfrac{24 \div 3}{24 \div 3} = \dfrac{2}{3} * \dfrac{8}{8} = \dfrac{16}{24}.[/MATH]
 
JeffM said:
You are confusing yourself by mixing together instructions in English with operations in math.

You seem to have been taught the following procedure.

To find the equivalent fraction to 2/3 with a denominator of 24:
(1) Divide 24 by 3, getting 8;
(2) Multiply 2 by 8, getting 16;
(3) Put 16 over 24 and get the following: [MATH]\dfrac{16}{24} = \dfrac{2}{3}.[/MATH]
You want to know why it works from first principles.

[MATH]a \ne 0 \implies \dfrac{a}{a} = 1.[/MATH]
[MATH]b = b * 1.[/MATH]
Two basic principles of arithmetic.

What the English instructions do is to summarize the following sequence of arithmetic operations.

[MATH]\dfrac{2}{3} = \dfrac{2}{3} * 1 = \dfrac{2}{3} * \dfrac{24 \div 3}{24 \div 3} = \dfrac{2}{3} * \dfrac{8}{8} = \dfrac{16}{24}.[/MATH]
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I want to thank you
JeffM said:
You are confusing yourself by mixing together instructions in English with operations in math.

You seem to have been taught the following procedure.

To find the equivalent fraction to 2/3 with a denominator of 24:
(1) Divide 24 by 3, getting 8;
(2) Multiply 2 by 8, getting 16;
(3) Put 16 over 24 and get the following: [MATH]\dfrac{16}{24} = \dfrac{2}{3}.[/MATH]
You want to know why it works from first principles.

[MATH]a \ne 0 \implies \dfrac{a}{a} = 1.[/MATH]
[MATH]b = b * 1.[/MATH]
Two basic principles of arithmetic.

What the English instructions do is to summarize the following sequence of arithmetic operations.

[MATH]\dfrac{2}{3} = \dfrac{2}{3} * 1 = \dfrac{2}{3} * \dfrac{24 \div 3}{24 \div 3} = \dfrac{2}{3} * \dfrac{8}{8} = \dfrac{16}{24}.[/MATH]
Click to expand...
I want to thank you for your reply. I understand the arithmetic involved with your explanation, but it does notexactly answer my question.

Let me use a different approach. Suppose that I want to finance something for $200. The finance charge is $50 for the $200. I can express it as $50/$200. Now how much is the finance charge for $100? I can determine the amount by the following: $50/$200 * $100. Now $200 goes or divides into $100 1/2 times. Now I must use the $50 in the calculation. So, I multiply it by 1/2 to get the amount of $25. What I am seeing here is something that is proportional. That is the best way that I can explain it. I can do the math, but why it works out is more difficult to see or understand.
 
Explain this! said:
I want to thank you

I want to thank you for your reply. I understand the arithmetic involved with your explanation, but it does notexactly answer my question.

Let me use a different approach. Suppose that I want to finance something for $200. The finance charge is $50 for the $200. I can express it as $50/$200. Now how much is the finance charge for $100? I can determine the amount by the following: $50/$200 * $100. Now $200 goes or divides into $100 1/2 times. Now I must use the $50 in the calculation. So, I multiply it by 1/2 to get the amount of $25. What I am seeing here is something that is proportional. That is the best way that I can explain it. I can do the math, but why it works out is more difficult to see or understand.
Click to expand...
It is the exact same process. You want a fraction (or proportion) with a denominator of 100 equivalent to 50/200.

[MATH]\dfrac{50}{200} = \dfrac{50}{200} * 1.[/MATH] No problem there.

You say [MATH]100 \div 200 = 1/2. [/MATH] Absolutely correct.

And I say [MATH]\dfrac{1/2}{1/2} = 1.[/MATH] Any argument there?

[MATH]\therefore \dfrac{50}{200} = \dfrac{50}{200} * 1 = \dfrac{50}{200} * \dfrac{1/2}{1/2}.[/MATH]
Now we get to your first division step.

[MATH]\therefore \dfrac{50}{200} = \dfrac{50}{200} * \dfrac{1/2}{1/2} = \dfrac{50}{200 * \frac{1}{2}} * \dfrac{1/2}{1} = \dfrac{50}{200 \div 2} * \dfrac{1/2}{1} = \dfrac{50}{100} * \dfrac{1/2}{1}.[/MATH]
Now we get to your second division step.

[MATH]\therefore \dfrac{50}{200} = \dfrac{50}{100} * \dfrac{1/2}{1} = \dfrac{50 * 1/2}{100} = \dfrac{50 \div 2}{100} = \dfrac{25}{100}.[/MATH]
Your algorithm in English describes a valid sequence of arithmetic steps. (Of course lex is correct that algebra and cross-multiplication are quicker and just as valid, but that is helpful only if you know algebra.)
 
JeffM said:
It is the exact same process. You want a fraction (or proportion) with a denominator of 100 equivalent to 50/200.

[MATH]\dfrac{50}{200} = \dfrac{50}{200} * 1.[/MATH] No problem there.

You say [MATH]100 \div 200 = 1/2. [/MATH] Absolutely correct.

And I say [MATH]\dfrac{1/2}{1/2} = 1.[/MATH] Any argument there?

[MATH]\therefore \dfrac{50}{200} = \dfrac{50}{200} * 1 = \dfrac{50}{200} * \dfrac{1/2}{1/2}.[/MATH]
Now we get to your first division step.

[MATH]\therefore \dfrac{50}{200} = \dfrac{50}{200} * \dfrac{1/2}{1/2} = \dfrac{50}{200 * \frac{1}{2}} * \dfrac{1/2}{1} = \dfrac{50}{200 \div 2} * \dfrac{1/2}{1} = \dfrac{50}{100} * \dfrac{1/2}{1}.[/MATH]
Now we get to your second division step.

[MATH]\therefore \dfrac{50}{200} = \dfrac{50}{100} * \dfrac{1/2}{1} = \dfrac{50 * 1/2}{100} = \dfrac{50 \div 2}{100} = \dfrac{25}{100}.[/MATH]
Your algorithm in English describes a valid sequence of arithmetic steps. (Of course lex is correct that algebra and cross-multiplication are quicker and just as valid, but that is helpful only if you know algebra.)
Click to expand...

Yes, I understand your process. I want to thank you for your assistance and efforts!
 
What you are doing is what I call checking a division problem.

For example, 10/2 = 5 since 2*5 = 10.
15/3 = 5, since 3*5 = 15.

x/3 = 5 since 3*5 = x. So x=15

x/7 = 3 since 7*3 = x. That is, x= 21

Just like 4 is one number, 2/3 is one number. 1/3, 5/13, 4/5,... are all one number.

So when you have 2/3 = ?/6, then 6*(2/3) = ? or ? = 4

If you have 7/9 = ?/27, then 27*(7/9) = ?. So ? =21
 
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