# Comparing fractions, explaining differences

• Aug 3rd 2006, 06:14 PM
Euclid Alexandria
Comparing fractions, explaining differences
From the book: "Which fraction is closest to 0? Explain your answer."

a. $\displaystyle \frac{1}{2}$
b. $\displaystyle \frac{1}{3}$
c. $\displaystyle \frac{1}{4}$
d. $\displaystyle \frac{1}{5}$

Answer = d. 1 part out of 5 parts is the smallest fraction given.

Is the explanation of my answer sufficient, or is there a clearer way for me to explain what makes my answer the smallest fraction given?.
• Aug 3rd 2006, 06:30 PM
ThePerfectHacker
Quote:

Originally Posted by Euclid Alexandria
From the book: "Which fraction is closest to 0? Explain your answer."

a. $\displaystyle \frac{1}{2}$
b. $\displaystyle \frac{1}{3}$
c. $\displaystyle \frac{1}{4}$
d. $\displaystyle \frac{1}{5}$

Answer = d. 1 part out of 5 parts is the smallest fraction given.

Is the explanation of my answer sufficient, or is there a clearer way for me to explain what makes my answer the smallest fraction given?

Since these numbers are all positive all you need is to find the smallest number.
Think of this.
If $\displaystyle x>y$ (x is larger than y)
then,
$\displaystyle \frac{1}{x} < \frac{1}{y}$ (less than), why?
Simple, what does $\displaystyle \frac{1}{x}$ mean?
Take a whole part (a whole is 1) and divide it into $\displaystyle x$ equal part. But since "x" is larger than "y" then the parts of $\displaystyle \frac{1}{x}$ are smaller than the parts of $\displaystyle \frac{1}{y}$. So in your problem find the number with the Largest denominator. Which is, $\displaystyle \frac{1}{5}$
• Aug 3rd 2006, 07:22 PM
Quick
Quote:

Originally Posted by ThePerfectHacker
Since these numbers are all positive all you need is to find the smallest number.
Think of this.
If $\displaystyle x>y$ (x is larger than y)
then,
$\displaystyle \frac{1}{x} < \frac{1}{y}$ (less than)

I find this statement interesting hacker, does it mean that when you give both sides a negative exponent than the sign must be flipped? (obviously you can look at it logically and say it's correct, but I find it interesting how mathematicians have to use math all the time to prove these things)
• Aug 3rd 2006, 08:02 PM
ThePerfectHacker
Quote:

Originally Posted by Quick
(obviously you can look at it logically and say it's correct,

Quick, intuition (using logic) is not always true. You many times miss something and you are wrong. This is why mathematicians do everything strictly because it avoids missing something.

Formally, $\displaystyle a>b$ means, $\displaystyle a-b\in \mathbb{R}^+$ (meaning it is postive real number. [I know what you are thinking how do you define positive? There is a way but we will ignore that] )

So we are given that $\displaystyle x>y$ and we need to prove that, $\displaystyle \frac{1}{x} < \frac{1}{y}$ (since we do not want black holes we of course have $\displaystyle x,y\not = 0$).
But by the above defintion we need to prove that,
$\displaystyle \frac{1}{x}-\frac{1}{y}<0$
Thus,
$\displaystyle \frac{y-x}{xy}<0$ (1)
But,
$\displaystyle y-x \mbox{ is negative }$ because $\displaystyle x-y>0$ (given conditions)

Now, note the numerator of (1) is a negative number, however the denominator depends on the the signs of $\displaystyle x,y$. If both $\displaystyle x,y$ have the same same their product is positive otherwise not. Therefore, we have a negative number ($\displaystyle y-x$) divided by a positive number ($\displaystyle xy$) is a negative number. Therefore, we proved $\displaystyle \frac{1}{x} < \frac{1}{y}$ whenever they both got the same sign

Quote:

Originally Posted by Quick
but I find it interesting how mathematicians have to use math all the time to prove these things)

Many people (usually physisisists) do not see the beauty of math and thing proving the fact,
$\displaystyle x>y \mbox{ and } y>z \mbox{ then } x>z$ useless, but not mathematicians because they avoid human logic. And thus never in fear of being wrong in some unique case
• Aug 16th 2006, 06:33 PM
Euclid Alexandria
Thanks Hacker
After reading over that a couple times, it makes sense. Somewhat related to Quick's comment-- it's strange how I had to learn how to explain something logically that I already knew intuitively. Anyway here's my lengthier (but clearer) answer:

Answer = d. 1 part out of 5 parts is the closest to 0. Imagine four pies. $\displaystyle \frac{1}{2}$ is 1 piece of a pie cut equally into two pieces, and $\displaystyle \frac{1}{3}$ is 1 piece of a pie cut into three smaller equal pieces, and $\displaystyle \frac{1}{4}$ is 1 piece of a pie cut into four even smaller equal pieces, and $\displaystyle \frac{1}{5}$ is 1 piece of a pie cut into five equal pieces smaller than in all the other fractions. Cutting this last pie into smaller pieces makes the fraction (the piece) smaller, which in turn means this fraction must be the closest to 0. Sound good? (The answer, not the pie.)
• Aug 16th 2006, 07:05 PM
Quick
It does, especially if you like analogies.

but why not give all the fractions a common denominator and then compare them?

such as:

$\displaystyle 0<\frac{24}{120}<\frac{30}{120}<\frac{40}{120}<\fr ac{60}{120}\quad\Longrightarrow\quad 0<\frac{1}{5}<\frac{1}{4}<\frac{1}{3}<\frac{1}{2}$

that seems to make it perfectly clear which ones closest to zero...
• Sep 1st 2006, 09:51 PM
Euclid Alexandria
I'm having a hard time understanding this explanation.
$\displaystyle \frac{14}{120} = \frac{7}{60}$
I was expecting it to equal $\displaystyle \frac{1}{5}$
(I would have seen how it relates to my fractions if that were the case.)
Could you explain further?
• Sep 1st 2006, 11:42 PM
Random333
Quote:

Originally Posted by Euclid Alexandria
I'm having a hard time understanding this explanation.
$\displaystyle \frac{14}{120} = \frac{7}{60}$
I was expecting it to equal $\displaystyle \frac{1}{5}$
(I would have seen how it relates to my fractions if that were the case.)
Could you explain further?

If it was $\displaystyle \frac{14}{70}$ then it would be $\displaystyle \frac{1}{5}$ because 70 divided by 14 equals 5.

Alternatively if the question was $\displaystyle \frac{24}{120}$ then the answer would be $\displaystyle \frac{1}{5}$ because 120 divided by 24 equals 5.

In $\displaystyle \frac{14}{120}$ the Highest Common Factor(HCF) is 2, therefore,
$\displaystyle \frac{14}{120} = \frac{7}{60}$
• Sep 2nd 2006, 02:49 AM
Quick
Sorry :o
As random said, I should have wrote $\displaystyle \frac{24}{120}=\frac{1}{5}$
• Oct 20th 2006, 02:16 AM
Euclid Alexandria
Thinking process?
Quote:

Originally Posted by Quick
It does, especially if you like analogies.

but why not give all the fractions a common denominator and then compare them?

such as:

$\displaystyle 0<\frac{14}{120}<\frac{30}{120}<\frac{40}{120}<\fr ac{60}{120}\quad\Longrightarrow\quad 0<\frac{1}{5}<\frac{1}{4}<\frac{1}{3}<\frac{1}{2}$

that seems to make it perfectly clear which ones closest to zero...

Hi, Quick, I was away from the forums (and math classes) for a while, and this is one of the posts I still have questions about.

After Random's correction, I can see that your fractions (with a common denominator) are relevant to my fractions. I checked the rest of your fractions and they're also clearly relevant. They also make a briefer explanation than my analogy (I'm an intuitive thinker and a writer, so I often think in analogous terms, which obviously aren't the briefest terms in this case).

However, if I were presented with a similar question as the one posed in my original post, I would not be able to easily recreate an explanation similar to yours. This is because I can't see how you came to the conclusion that using 120 for a common denominator would be a good idea, nor can I see how you decided on what numerators to use (without the long and laborious process of filling a page with trial-and-error solutions, in which case an analogous answer would be much simpler in practice).

Could you please explain the thinking process by which you came up with your explanation of why $\displaystyle \frac{1}{5}$ is the smallest fraction?
• Oct 20th 2006, 02:59 AM
AfterShock
Quote:

Originally Posted by Euclid Alexandria
...I can't see how you came to the conclusion that using 120 for a common denominator would be a good idea, nor can I see how you decided on what numerators to use (without the long and laborious process of filling a page with trial-and-error solutions, in which case an analogous answer would be much simpler in practice).

Could you please explain the thinking process by which you came up with your explanation of why $\displaystyle \frac{1}{5}$ is the smallest fraction?

He used 120 for a common denominator; it would have been better to use 60 as a common denominator (but both obviously work, as they are multiples of each other). In order to find a common denominator, you try and find a number that ALL of the numbers go into that number. The smallest in this case is 60.

Take 1/5 for instance.

What is 1/5 of 60.

Isn't it just 60/5? So it is 12.

30.

Similarly, we can find 1/3 and 1/4.

60/3 = 20

1/4 = 15.

Which is the closest to 0? As we can see, 1/5 is, as it "goes into" 60 exactly 12 times. That is, 12/60 is equivalent to 1/5.

Therefore, 1/5 is the fraction that is closest to zero.
• Oct 20th 2006, 06:03 AM
topsquark
Quote:

Originally Posted by Euclid Alexandria
Hi, Quick, I was away from the forums (and math classes) for a while, and this is one of the posts I still have questions about.

After Random's correction, I can see that your fractions (with a common denominator) are relevant to my fractions. I checked the rest of your fractions and they're also clearly relevant. They also make a briefer explanation than my analogy (I'm an intuitive thinker and a writer, so I often think in analogous terms, which obviously aren't the briefest terms in this case).

However, if I were presented with a similar question as the one posed in my original post, I would not be able to easily recreate an explanation similar to yours. This is because I can't see how you came to the conclusion that using 120 for a common denominator would be a good idea, nor can I see how you decided on what numerators to use (without the long and laborious process of filling a page with trial-and-error solutions, in which case an analogous answer would be much simpler in practice).

Could you please explain the thinking process by which you came up with your explanation of why $\displaystyle \frac{1}{5}$ is the smallest fraction?

When given a list of fractions, $\displaystyle \frac{a}{b}, \, \frac{c}{d},...$ a useful common denominator is simply the product of all the denominators. In your example you had the denominators 2, 3, 4, and 5. So a useful common denominator would be 2*3*4*5 = 120. We can express all of your original fractions as fractions with a denominator of 120. To get the new numerators:

$\displaystyle \frac{1}{2} = \frac{x}{120}$

Now solve for x:
$\displaystyle x = \frac{1}{2} \times 120 = 60$

For $\displaystyle \frac{1}{3}$ we get $\displaystyle \frac{1}{3} = \frac{40}{120}$

etc.

If you have heard of it (I don't know what grade you are in) there is another useful common denominator you can find called the "least common multiple." This is the smallest number that is the multiple of a list of numbers. In this case we are looking for the least common multiple of 2, 3, 4, and 5. It turns out to be 60. No smaller number is a multiple of all of 2, 3, 4, and 5. I'll show you how to do this if you wish, but if you haven't been introduced to the term it might be better to simply wait until your classes catch up with it.

-Dan