# Writing a Fraction in Simplest Form

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Jun 22nd 2006, 07:00 PM
Euclid Alexandria
Writing a Fraction in Simplest Form
Say we are dividing out the common factors in $\frac{7 \cdot a \cdot a \cdot a}{7 \cdot 8 \cdot a \cdot a}$

Is an answer of $\frac{1a}{8}$ just as valid and correct as $\frac{a}{8}$?
• Jun 22nd 2006, 07:58 PM
Soroban
Hello, Euclid Alexandria!

Quote:

Say we are dividing out the common factors in $\frac{7 \cdot a \cdot a \cdot a}{7 \cdot 8 \cdot a \cdot a}$

Is an answer of $\frac{1a}{8}$ just as valid and correct as $\frac{a}{8}$?
Yes, it is . . . but a coefficient of $1$ is usually omitted.

Similarly: . $\frac{7\cdot a\cdot a\cdot a}{a\cdot a}$ is equal to $\frac{7a}{1}$,
. . but a denominator of $1$ is usually omitted.
• Jun 23rd 2006, 03:33 AM
Quick
Quote:

Originally Posted by Euclid Alexandria
Say we are dividing out the common factors in $\frac{7 \cdot a \cdot a \cdot a}{7 \cdot 8 \cdot a \cdot a}$

Is an answer of $\frac{1a}{8}$ just as valid and correct as $\frac{a}{8}$?

I've had some teachers that marked me down for doing something like that :eek: so if you're giving the answers to a test, it is better to get rid of the 1.
• Jun 23rd 2006, 05:51 PM
Euclid Alexandria
Thanks guys, I'll get in the habit of omitting the 1 just in case.

Oh hey look, I just became a senior member.
• Jun 23rd 2006, 07:44 PM
ticbol
Quote:

Originally Posted by Euclid Alexandria
Thanks guys, I'll get in the habit of omitting the 1 just in case.

Oh hey look, I just became a senior member.

That means you're old now. You're entitled to some discounts now. You can be given incomplete answers now.
• Jun 24th 2006, 12:59 PM
Euclid Alexandria
Quote:

Originally Posted by ticbol
That means you're old now. You're entitled to some discounts now. You can be given incomplete answers now.

I'm glad to hear $\frac{3 \cdot a \cdot a}{9 \cdot a \cdot a}$ of that!
• Jun 24th 2006, 06:01 PM
malaygoel
Writing fraction in simplest form
When we write a fraction in the simplest form, we cancel out what is common in the numerator and denominator. This is what a teacher tells us in a class. After this the practice problem was given:
Simplify $\frac{16}{64}$
One student cancelled out the $6$(common in numerator and denominator) and got the answer $\frac{1}{4}$, which was suprisingly the right answer. How many more fractions are there like this?

KeepSmiling
Malay
• Jun 24th 2006, 09:21 PM
Soroban
Hello, Malay!

Besides $\frac{1\!\!\!\not{6}}{\not{6}4} = \frac{1}{4}$, we have:

. . $\frac{2\!\!\!\not{6}}{\not{6}5} = \frac{2}{5}$

. . $\frac{1\!\!\!\not{9}}{\not{9}5} = \frac{1}{5}$

. . $\frac{4\!\!\!\not{9}}{\not{9}8} = \frac{4}{8} = \frac{1}{2}$
• Jun 26th 2006, 06:41 PM
Euclid Alexandria
Quote:

Originally Posted by Soroban
Hello, Malay!

Besides $\frac{1\!\!\!\not{6}}{\not{6}4} = \frac{1}{4}$, we have:

. . $\frac{2\!\!\!\not{6}}{\not{6}5} = \frac{2}{5}$

. . $\frac{1\!\!\!\not{9}}{\not{9}5} = \frac{1}{5}$

. . $\frac{4\!\!\!\not{9}}{\not{9}8} = \frac{4}{8} = \frac{1}{2}$

Interesting...are these all the two-digit fractions that are like this? Did you know this intuitively or did you use a method to figure it out?
• Jun 26th 2006, 09:11 PM
Soroban
Hello, Euclid Alexandria!

Quote:

Are these all the two-digit fractions that are like this?
Did you know this intuitively or did you use a method to figure it out?

I had seen these listed a book many years ago.
But you can derive your own formula for finding these "jokes".

We want digits $a,b,c$ so that: . $\frac{10a + b}{10b + c} \:= \:\frac{a}{c}$

Solve for $c:\;\;c\:=\:\frac{10ab}{9a + b}$

Now try all digits for $a$ and $b$ which make $c$ a digit.

You'll find that the list I gave is complete.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

One more joke:

Reduce: . $\frac{(1 + x)^2}{1 - x^2}$

Answer: . $\frac{(1 + x)^{\not{2}}} {1 - x^{\not{2}}} \;=\;\frac{1 + x}{1 - x}$

• Jun 27th 2006, 06:43 PM
Euclid Alexandria
Quote:

Originally Posted by Soroban
I had seen these listed a book many years ago.
But you can derive your own formula for finding these "jokes".

We want digits $a,b,c$ so that: . $\frac{10a + b}{10b + c} \:= \:\frac{a}{c}$

Solve for $c:\;\;c\:=\:\frac{10ab}{9a + b}$

Now try all digits for $a$ and $b$ which make $c$ a digit.

You'll find that the list I gave is complete.

I am having difficulty following the example and reproducing the results. Here are the steps I took to reproduce the example:

$
\frac{1\!\!\!\not{6}}{\not{6}4} = \frac{1}{4}
$

Let a = 1, b = 6, and c = 4.

$\frac{10a + b}{10b + c} = \frac{a}{c}$

$c = \frac{10ab}{9a + b}$

$\frac{10 \cdot 1 \cdot 6}{10 \cdot 6 + 4}$

$= \frac{16}{20}$

$= \frac{1 \cdot \not2 \cdot 3}{5 \cdot \not2 \cdot 2}$

$= \frac{3}{10}$
• Jun 28th 2006, 04:59 AM
Soroban
Hello, Euclid Alexandria!

A few really gruesome errors . . .

Quote:

I am having difficulty following the example and reproducing the results. . . . No wonder!

Here are the steps I took to reproduce the example:

$\frac{1\!\!\!\not{6}}{\not{6}4} = \frac{1}{4}$

Let a = 1, b = 6, and c = 4.

$\frac{10a + b}{10b + c} = \frac{a}{c}$

$c = \frac{10ab}{9a + b}$

$= \frac{10 \cdot 1 \cdot 6}{10 \cdot 6 + 4}$
. . . . ↑? . .↑?
.ten? . . b = 6
. . . . a = 1

. 10·1·6 = 60
. . . ↓?
$= \frac{16}{20}$
. . . ↑?
. 10·6 + 4 = 64

. What is this? . 16 = 2·2·2·2
. . . . .
$= \frac{1 \cdot \not2 \cdot 3}{5 \cdot \not2 \cdot 2}$

$= \frac{3}{10}$
Can you possibly make any more mistakes?
• Jun 28th 2006, 08:51 AM
topsquark
Quote:

Originally Posted by Soroban
Hello, Euclid Alexandria!

A few really gruesome errors . . .

Can you possibly make any more mistakes?

It's mathematically incorrect, but I thought she had done it on purpose. If so, I found it rather humerous. Here's why:

$\frac{10 \cdot 1 \cdot 6}{10 \cdot 6 +4}$
Well, 10 x 1 = 10 and thus 10 x 1 x 6 = 10 x 6 = 1(0+6)=16.
6 + 4 = 10 so 10 * 6 + 4 = 10 * 10 = (1+1)0 = 20.

$\frac{16}{20}$
Now, 16 = 1 (6) = 1 * (2 * 3)
20 = 5 * 2 * 2

$\frac{1 \cdot 2 \cdot 3}{5 \cdot 2 \cdot 2}=\frac{3}{10}$

All of the Math mistakes are common (at least I've seen alot of them) mistakes similar to the cancelling joke that started this whole conversation.

(At least I HOPE she was making a joke out of it!!)

-Dan
• Jun 28th 2006, 06:53 PM
Euclid Alexandria
Yes, it is possible to make more mistakes in this instance. :eek:

:) Thoughtful pause :)

Here we are attempting to reproduce $\frac{1\!\!\!\not{6}}{\not{6}4} = \frac{1}{4}$ as before, except with a newly revised method.

Let a = 1, b = 6, and c = 4

$
\frac{10a + b}{10b + c} = \frac{a}{c}
$

$
c = \frac{10ab}{9a + b}
$

$= \frac{10 \cdot 1 \cdot 6}{9 \cdot 1 + 6}$
. . .. ...
.nine! . b = 6
. . . . a = 1

$= \frac{60}{15} = \frac{2 \cdot 2 \cdot \not3 \cdot \not5}{\not 3 \cdot \not5}$

$= \frac{4}{1} = 4$
• Jun 30th 2006, 06:44 PM
Euclid Alexandria
Here's another try at using that formula (if that's the correct terminology) to reproduce $\frac{1\!\!\!\not{6}}{\not{6}4} = \frac{1}{4}$

Let a = 1, b = 6, and c = 4

$\frac{10a + b}{10b + c} = \frac{a}{c}$

$= \frac{10 \cdot 1 + 6}{10 \cdot 6 + 4} = \frac{16}{64}$

(Now it appears that a = 16 and c = 64)

$c = \frac{10ab}{9a+b}$

$= \frac{10 \cdot 16 \cdot 6}{9 \cdot 16 + 6} = \frac{960}{150}$

$= \frac{\not2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot \not3 \cdot \not5}{\not2 \cdot \not3 \cdot \not5 \cdot 5}$

$= \frac{32}{5}$
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last