Subcategories

• Module 1 Day 1 Challenge Part 1

  Decimal to Fraction Conversion, Repeating Decimals

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• Module 1 Day 1 Challenge Part 2

  Repeating Decimals Like 0.3333...

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• Module 1 Day 1 Challenge Part 3

  Improper Fractions, Multiplying and Simplifying Fractions

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  @funnyorangutan-1 Thanks for your very polite and curious responses. 🙂 The first question about why \(\frac{1}{3} = 0.\overline{333}\ldots \) is quite long, so if you don't mind, I'll put that in it's own post, here. https://forum.poshenloh.com/topic/545/why-is-0-3333 ⬅ ⬅ ⬅ Answer to why \(\frac{1}{3} = 0.\overline{333}\ldots \)

  For your other question, about why \( 0.\overline{222}\ldots = \frac{2}{9} \text{ and } 0.\overline{444}\ldots = \frac{4}{9}\) and so on, maybe this diagram will help a bit:

  M1-forum-fractions-with-denominator-9.png

  You can treat \(0.\overline{111}\ldots\) as a starting point from which you can multiply by any factor in order to get any fraction of the form \(\frac{n}{9},\) where \(n\) is an integer.

  "Oh, but didn't you leave out the \(\frac{3}{9} \text{ and } \frac{6}{9}?\)"

  Very perceptive question! 🙂

  I left those out because \(\frac{3}{9}\) and \(\frac{6}{9}\) aren't in simplified form. You wouldn't put "\(\frac{3}{9}\)" on a test as an answer. Instead, you would put \(\frac{1}{3}.\) And you wouldn't write "\(\frac{6}{9}\)" on a test as an answer either -- you would write \(\frac{2}{3}.\) So even though the pattern holds for multiplying \(0.\overline{111}\ldots\) by both \(3\) and \(6,\) I left it out, but it is very nice to know that everyone holds across math, and that everything works as it should, because

  $$ 0.\overline{333}\ldots = \frac{3}{9} = \frac{1}{3} $$

  Great, math is consistent! 🙂

  When the world seems like it's failing you and there's no end in sight......... there's always math! 🎉 🎉 🎉

• Module 1 Day 1 Your Turn Part 1

  Fraction to Decimal Conversion

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  D

  @the-blade-dancer The second step looks quite different from the first step! It's because for the first two fractions, we have multiplied the top and bottom by a factor.

  For the first fraction, we have multiplied the top and bottom by \( n + 1.\)

  \(\frac{1}{\textcolor{red}{n}} = \frac{n + 1}{\textcolor{red}{n} (n + 1) } \)

  For the second fraction, we have multiplied the top and bottom by \(n.\)

  \(\frac{1}{\textcolor{red}{n+1}} = \frac{n}{\textcolor{red}{(n +1)} n } \)

  For the third fraction, we've done something different: we have factored the bottom. (Factor just means to write as a product of two terms, like \( 12 = 3 \times 4.\) )

  \( - \frac{1}{n^2 + n} = - \frac{1}{n(n+1)} \)

  Why are we trying to do this? To make the bottoms of the fractions all equal to \( n(n+1).\) Then we can add the fractions together very easily, and luckily we have some terms cancel out, so we get \(0\) on the top, which is a little fortunate!

  \( \frac{n + 1 - n - 1}{n(n+1)} \)

  \( = \frac{0}{n(n+1)} = \boxed{0}.\)

  🙂

  high-five.png

• Module 1 Day 1 Your Turn Part 2

  Common Denominators

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  E

  @inventiveantelope Doesn't really matter.