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text: '[summary: This page consists of exercises designed to help you get to grips with the addition of rational numbers.]\n\nThis page consists of exercises designed to help you get to grips with the addition of rational numbers.\n\n# Core exercises\n\n## First exercise: $\\frac{1}{10} + \\frac{1}{5}$\n\n%%hidden(Show some possible solutions):\nUsing the instant rule from the text (which is actually a bit unwieldy here): $$\\frac{1}{10} + \\frac{1}{5} = \\frac{1 \\times 5 + 10 \\times 1}{10 \\times 5} = \\frac{5+10}{50} = \\frac{15}{50}$$\nNotice that this can actually be made simpler: it is the same thing as $\\frac{3}{10}$, because when we take $\\frac{3}{10}$ and split each $\\frac{1}{10}$-block into five equal pieces, we obtain $15$ copies of the $\\frac{1}{50}$-block.\n\nAlternatively, one could spot that $\\frac{1}{10}$-blocks actually can already be used to make $\\frac{1}{5}$-blocks: $\\frac{1}{5} = \\frac{2}{10}$.\nTherefore we're actually trying to find $\\frac{1}{10} + \\frac{2}{10}$, which is easy: it's $\\frac{3}{10}$.\n%%\n\n## Second exercise: $\\frac{1}{15} + \\frac{1}{10}$\n\n%%hidden(Show some possible solutions):\nUsing the instant rule from the text: $$\\frac{1}{10} + \\frac{1}{15} = \\frac{1 \\times 15 + 10 \\times 1}{10 \\times 15} = \\frac{25}{150} = \\frac{1}{6}$$\n\nWere you expecting a big denominator, or at least a multiple of 5? From this example, we can see that the final answer of a rational addition problem can have a denominator which doesn't even seem related to the others.\n\nAlternatively, one could spot that $\\frac{1}{30}$-blocks will make up $\\frac{1}{10}$- and $\\frac{1}{15}$-blocks.\nThen we are actually trying to find $\\frac{3}{30} + \\frac{2}{30} = \\frac{5}{30}$; it's a bit easier to see that $\\frac{5}{30} = \\frac{1}{6}$ than it is to see that $\\frac{25}{150} = \\frac{1}{6}$.\n%%\n\n## Third exercise: $\\frac{1}{10} + \\frac{1}{15}$\nYou might find this exercise a little familiarâ€¦\n\n%%hidden(Show possible solution):\nUsing the instant rule from the text: $$\\frac{1}{15} + \\frac{1}{10} = \\frac{1 \\times 10 + 15 \\times 1}{15 \\times 10} = \\frac{25}{150} = \\frac{1}{6}$$\n\nYou may notice that we've basically done the same calculations as in the second exercise.\nIn fact, addition doesn't care which way round the numbers go: $\\frac{1}{10} + \\frac{1}{15} = \\frac{1}{15} + \\frac{1}{10}$, even if we don't already know that that number is $\\frac{1}{6}$.\n\nThis is intuitive from the fact that addition is the idea of "place the apples next to each other and count up the total": just putting the apples down in a different order doesn't change the total amount of apple.\n%%\n\n## Fourth exercise: $\\frac{0}{5} + \\frac{2}{5}$\n\n%%hidden(Show possible solution):\nNotice that both the denominators are the same (namely $5$), so we can just combine the $\\frac{1}{5}$-sized pieces straight away.\nWe have $0$ pieces and $2$ pieces, so the total is $2$ pieces.\n\nThat is, the answer is $\\frac{2}{5}$.\n%%\n\n## Fifth exercise: $\\frac{0}{7} + \\frac{2}{5}$\n\n%%hidden(Show possible solution):\nIf you spot that this is "no $\\frac{1}{7}$-pieces" next to "two $\\frac{1}{5}$-pieces", then you might just immediately write down that the answer is $\\frac{2}{5}$ because there aren't any $\\frac{1}{7}$-pieces to change the answer; and you'd be correct.\n\nTo use the instant rule from the text: $$\\frac{0}{7} + \\frac{2}{5} = \\frac{0 \\times 5 + 2 \\times 7}{5 \\times 7} = \\frac{0 + 14}{35} = \\frac{14}{35}$$\n\nBut that is the same as $\\frac{2}{5}$ (simply expressed with each $\\frac{1}{5}$-piece subdivided further into sevenths).\n%%\n\n# Extension exercises\n\nThese exercises are meant to be harder and to stretch your conceptual understanding.\nGive them a proper go, but don't worry too much if you don't get the same answers as me.\nMine are, in a technical sense, "right", but no matter what you end up with, you will derive a lot of benefit from trying to work out what the answers are yourself without having been told exactly how.\nThe learning of mathematics is much more about thinking and understanding (usually guided by examples) than it is about just repeatedly carrying out calculations.\n\n## First extension exercise: $\\frac{1}{5} + \\frac{-1}{10}$\nYes, it is possible to add a negative number of chunks.\nTry using the instant rule and see what happens.\n\n%%hidden(Show possible solution):\nUsing the instant rule from the text: $$\\frac{1}{15} + \\frac{-1}{10} = \\frac{1 \\times 10 + 15 \\times (-1)}{15 \\times 10} = \\frac{10 - 15}{150} = \\frac{-5}{150} = \\frac{-1}{30}$$\n\nWhat has happened here? What have we "really done" with our chunks of apple?\nHave a think; we'll see a lot more of this when we get to [56x subtraction].\n%%\n\n## Second extension exercise: what rational number must we add to $\\frac{7}{8}$ to obtain $\\frac{13}{8}$?\n\nYou're looking for an answer that looks like $\\frac{a}{b}$ where there are integers in place of $a$ and $b$.\n\n%%hidden(Show some possible solutions):\n\nSomething you can do to make this question easier is to notice that both the numbers have the same chunk-size (namely $\\frac{1}{8}$), so we might try adding some number of $\\frac{1}{8}$-chunks.\nThen we're trying to get from $7$ chunks to $13$ chunks, so we need to add $6$ chunks.\n\nThat is, the final number is $\\frac{6}{8}$ (which is also $\\frac{3}{4}$).\n%%\n\n## Third extension exercise: what rational number must we add to $\\frac{7}{8}$ to obtain $\\frac{13}{7}$?\n\nYou're looking for an answer that looks like $\\frac{a}{b}$ where there are integers in place of $a$ and $b$.\n\n%%hidden(Show some possible solutions):\n\nNow, the numbers are no longer of the same chunk-size, so we should make it so that they are of the same size.\n\nThe chunk-size to use is $\\frac{1}{8 \\times 7} = \\frac{1}{56}$. The reason for this is the same as the reasoning we saw when working out how to add $\\frac{1}{8}$ and $\\frac{1}{7}$.\n\nThen the two numbers are $\\frac{7 \\times 7}{7 \\times 8} = \\frac{49}{56}$ and $\\frac{8 \\times 13}{8 \\times 7} = \\frac{104}{56}$, so the answer is that we need to get from $49$ to $104$; to do that, we need to add $55$ chunks of size $\\frac{1}{56}$, so the answer is $\\frac{55}{56}$.\n%%',
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