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

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

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- Jun 22nd 2006, 07:00 PMEuclid AlexandriaWriting a Fraction in Simplest Form
Say we are dividing out the common factors in $\displaystyle \frac{7 \cdot a \cdot a \cdot a}{7 \cdot 8 \cdot a \cdot a}$

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

Quote:

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

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

Similarly: .$\displaystyle \frac{7\cdot a\cdot a\cdot a}{a\cdot a}$ is equal to $\displaystyle \frac{7a}{1}$,

. . but a denominator of $\displaystyle 1$ is usually omitted. - Jun 23rd 2006, 03:33 AMQuickQuote:

Originally Posted by**Euclid Alexandria**

- Jun 23rd 2006, 05:51 PMEuclid 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 PMticbolQuote:

Originally Posted by**Euclid Alexandria**

- Jun 24th 2006, 12:59 PMEuclid AlexandriaQuote:

Originally Posted by**ticbol**

- Jun 24th 2006, 06:01 PMmalaygoelWriting 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 $\displaystyle \frac{16}{64}$

One student cancelled out the $\displaystyle 6$(common in numerator and denominator) and got the answer $\displaystyle \frac{1}{4}$, which was suprisingly the right answer. How many more fractions are there like this?

KeepSmiling

Malay - Jun 24th 2006, 09:21 PMSoroban
Hello, Malay!

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

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

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

. . $\displaystyle \frac{4\!\!\!\not{9}}{\not{9}8} = \frac{4}{8} = \frac{1}{2}$ - Jun 26th 2006, 06:41 PMEuclid AlexandriaQuote:

Originally Posted by**Soroban**

- Jun 26th 2006, 09:11 PMSoroban
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 $\displaystyle a,b,c$ so that: .$\displaystyle \frac{10a + b}{10b + c} \:= \:\frac{a}{c}$

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

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

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

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

One more joke:

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

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

- Jun 27th 2006, 06:43 PMEuclid AlexandriaQuote:

Originally Posted by**Soroban**

$\displaystyle

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

$

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

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

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

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

$\displaystyle = \frac{16}{20}$

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

$\displaystyle = \frac{3}{10}$ - Jun 28th 2006, 04:59 AMSoroban
Hello, Euclid Alexandria!

A few reallyerrors . . .*gruesome*

Quote:

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

Here are the steps I took to reproduce the example:

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

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

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

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

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

. . .**↑**.**↑?**. .**↑?**

.**ten?**.**↑**.**b = 6**

. . . .**a = 1**

.**10·1·6 = 60**

. . .**↓?**

$\displaystyle = \frac{16}{20}$

. . .**↑?**

.**10·6 + 4 = 64**

.*What is this?***. 16 = 2·2·2·2**

. . .**↓****. ↓ . ↓**

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

$\displaystyle = \frac{3}{10}$

- Jun 28th 2006, 08:51 AMtopsquarkQuote:

Originally Posted by**Soroban**

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

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.

$\displaystyle \frac{16}{20}$

Now, 16 = 1 (6) = 1 * (2 * 3)

20 = 5 * 2 * 2

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

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 PMEuclid Alexandria
Yes, it is possible to make more mistakes in this instance. :eek:

:)*Thoughtful pause*:)

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

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

$\displaystyle

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

$

$\displaystyle

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

$

$\displaystyle = \frac{10 \cdot 1 \cdot 6}{9 \cdot 1 + 6}$

. . ↑ .. ↑ ...↑

.**nine!**↑ .**b = 6**

. . . .**a = 1**

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

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

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

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

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

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

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

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

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

$\displaystyle = \frac{32}{5}$