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 $\displaystyle \frac{a}{8}$?
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 $\displaystyle \frac{a}{8}$?
Hello, Euclid Alexandria!
Yes, it is . . . but a coefficient of $\displaystyle 1$ is usually omitted.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.
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
Hello, Malay!
Besides $\displaystyle \frac{1\!\!\!\not{6}}{\not{6}4} = \frac{1}{4}$, we have:
. . $\displaystyle \frac{2\!\!\!\not{6}}{\not{6}5} = \frac{2}{5}$
. . $\displaystyle \frac{1\!\!\!\not{9}}{\not{9}5} = \frac{1}{5}$
. . $\displaystyle \frac{4\!\!\!\not{9}}{\not{9}8} = \frac{4}{8} = \frac{1}{2}$
Hello, Euclid Alexandria!
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: .$\displaystyle \frac{(1 + x)^{\not{2}}} {1 - x^{\not{2}}} \;=\;\frac{1 + x}{1 - x}$
I am having difficulty following the example and reproducing the results. Here are the steps I took to reproduce the example:Originally Posted by Soroban
$\displaystyle
\frac{1\!\!\!\not{6}}{\not{6}4} = \frac{1}{4}
$
Let a = 1, b = 6, and c = 4.
$\displaystyle \frac{10a + b}{10b + c} = \frac{a}{c}$
$\displaystyle c = \frac{10ab}{9a + b}$
$\displaystyle \frac{10 \cdot 1 \cdot 6}{10 \cdot 6 + 4}$
$\displaystyle = \frac{16}{20}$
$\displaystyle = \frac{1 \cdot \not2 \cdot 3}{5 \cdot \not2 \cdot 2}$
$\displaystyle = \frac{3}{10}$
Hello, Euclid Alexandria!
A few really gruesome errors . . .
Can you possibly make any more mistakes?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
. . .↓ . ↓ . ↓
$\displaystyle = \frac{1 \cdot \not2 \cdot 3}{5 \cdot \not2 \cdot 2}$
$\displaystyle = \frac{3}{10}$
It's mathematically incorrect, but I thought she had done it on purpose. If so, I found it rather humerous. Here's why: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
Yes, it is possible to make more mistakes in this instance.
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
$\displaystyle = \frac{60}{15} = \frac{2 \cdot 2 \cdot \not3 \cdot \not5}{\not 3 \cdot \not5}$
$\displaystyle = \frac{4}{1} = 4$
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)
$\displaystyle c = \frac{10ab}{9a+b}$
$\displaystyle = \frac{10 \cdot 16 \cdot 6}{9 \cdot 16 + 6} = \frac{960}{150}$
$\displaystyle = \frac{\not2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot \not3 \cdot \not5}{\not2 \cdot \not3 \cdot \not5 \cdot 5}$
$\displaystyle = \frac{32}{5}$