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Sorry for such an easy question but this has been confusing me like crazy. I have borrowed like normal and subtracted like every other problem, and the answer I get is -2.24. How is this wrong?

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    $\begingroup$ In order to answer, you need to present your work so that someone can find the error. $\endgroup$
    – user296602
    Jul 2, 2016 at 1:20
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    $\begingroup$ Think of it like this: What number do you need to add to $5.9$ to get $7.66$? $\endgroup$ Jul 2, 2016 at 1:21
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    $\begingroup$ I disagree a bit with the downvotes on this question. Usually a question formatted in this way should be downvoted, but OP is obviously both new to the community and not too familiar with math. The whole "downvote to let them vaguely know something's wrong" in this situation gives me a very StackOverflow-type vibe, and is really uninviting. $\endgroup$
    – galois
    Jul 2, 2016 at 1:55
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    $\begingroup$ @jaska I don't feel that "new to the community" is not an excuse for poor formatting and question style, and a good way to avoid downvotes - rather, it is a symptom of the overwhelming problem that people don't care about the quality of the post they make, so long as it gets a quick answer (as this one did four times...). $\endgroup$
    – user296602
    Jul 2, 2016 at 5:09
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    $\begingroup$ I answered the question because despite the OP's lack of exposition, it was painfully obvious what error they had almost certainly made. It occurs to me, however, that I think I've seen a question asked before about the same arithmetic error with different numbers. $\endgroup$
    – David K
    Jul 2, 2016 at 11:12

4 Answers 4

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$\require{cancel}$ \begin{array}{ll} &5.\quad\cancelto{8}{9}\,\cancelto{10}{0}\\ -\!\!\!&7.\quad6\quad6\\\hline -\!&2.\quad2\quad4 \end{array}

which is incorrect. So I don't think "borrowing" works when we subtract a "larger" (larger in magnitude) number from a smaller one (smaller in magnitude).

Instead, consider their magnitudes, that is, the distance each number is away from the $0$. Notice that the magnitude of $-7.66$ is larger than the magnitude of $5.90$, so we know that the final answer will be negative. So, we subtract the smaller magnitude number from the larger magnitude number and remember to force the final answer to be negative. Then we proceed as usual \begin{array}{r lll} &\cancelto{6}{7}.&\cancelto{16}{6}&6\\ -& 5\quad.&9&0\\\hline -&1\quad.&7&6 \end{array} which is correct, $-1.76$

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  • $\begingroup$ Thank you. I tried to give OP the benefit of the doubt. $\endgroup$
    – Em.
    Jul 2, 2016 at 2:03
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    $\begingroup$ What your original "wrong" subtraction neglected to show was that you are almost at the correct answer. What you needed to do was borrow a carry from the "nothing" to the left of the $5$ to make it a $15$. This would leave you with $8.24$ as a result, but to make up for the fact that you borrowed from an imaginary $10$, you need to undo that result by taking the negative of $10.00 - 8.24$, which is $-1.76$. $\endgroup$
    – Axoren
    Jul 2, 2016 at 2:26
  • $\begingroup$ @Axoren Yes, I see what you are saying but I was trying to strictly replicate (presumably) OP's approach. Also, I'm not sure that version of borrowing is typically taught. That would be useful though. $\endgroup$
    – Em.
    Jul 2, 2016 at 2:35
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    $\begingroup$ In fact, the $0.24$ resulting from the borrows is positive, assuming that the higher digits will have enough to borrow from. The result is really $-2 + 0.24=-1.76$ This is another way to get to the right answer. $\endgroup$ Jul 2, 2016 at 2:46
  • $\begingroup$ @RossMillikan right, the real problem is getting to the last step, and writing "-2" next to "0.24" and thinking that that makes -2.24. $\endgroup$
    – hobbs
    Jul 2, 2016 at 4:45
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If you consider the parts before and after the decimal point separately, you have this: \begin{align} 5.9 &= 5 + .9 \\ 7.66 &= 7 + .66 \\ \end{align} If you subtract the $7$ from the $5$ on one hand, and $0.66$ from $0.9$ on the other hand, you get \begin{align} 5 && .9\phantom{0} && \tag1\\ \underline{-7} && \underline{-.66} \tag2\\ -2 && .24 \tag3\\ \end{align} The key thing here is, what does the last line mean? On line $(1)$, the sum of the numbers in the two columns, $5$ and $0.9$, is one of the numbers we started with, $5.9$, while on line $(2)$, the sum of the numbers in the two columns, $-7$ and $-0.66$, is $-7.66$, corresponding to the operation of subtracting $7.66$. On line $(3)$, then, we must not subtract $0.24$ from $-2$ in order to obtain $-2.24$; rather, we must add the two numbers, like this: $$ (-2) + 0.24 = -(2 - 0.24) = -1.76. $$

In practice, this is still too confusing, and a better method is to rearrange the original subtraction so that when you do the digit-by-digit calculation you are always doing it with the larger-magnitude number on top. So we first write $$ 5.9 - 7.66 = - (-5.9 + 7.66) = -(7.66 - 5.9). $$ \begin{align} 7 &.66 \\ \underline{-5} & \underline{.9\phantom{0}} \\ 1 &.76 \end{align} Therefore $$ 5.9 - 7.66 = -(7.66 - 5.9) = -1.76. $$

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We have $5.9$. If you subtract one, you have $4.9$. Subtract $2$, you have $3.9$. Subtract $3$, $2.9$. Naively, one might expect that subtracting $7$ therefore gives you $-2.9$. This fails, however, because this "pattern" breaks down once you start going into the negatives. $0.9 - 1 = -0.1$. The pattern reemerges, so now $-1.1$; we have subtracted $7$ from $5.9$.

At this point, we deal with the decimal component. It's easy to see that the final answer is $-1.76$.

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5.9+0.1=6

6+1=7

7+0.66=7.66

0.1+1+0.66=1.76

why can we do it this way?

because 5.9-7.66=x can be written as

5.9-x=7.66

since 7.66>5.9, -x has to be positive, so x has to be negative, therefore the solution is -1.76.

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    $\begingroup$ 5.9-7.66=x cannot be written as 5.9+x=7.66 $\endgroup$ Jul 2, 2016 at 2:24

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