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My son was doing a math problem and getting the wrong result, so he asked me to help out. The problem was:

$4.5-8.2 = x$, solve for $x$

He was getting $-4.3$, where the correct answer is of course $-3.7$.

It turns out the reason he got the answer wrong is because he was doing the subtraction directly, and when he subtracted $.2$ from $.5$ he got $.3$. I showed him that he could multiply both sides by $-1$ and solve for $-x$, negate the answer and thus get the right result. Being the curious kid that he is, he asked why the two results weren't the same, and I couldn't give him an answer.

Why doesn't the direct subtraction work?

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    $\begingroup$ $$4.5-8.2 = \left(4+\frac{5}{10}\right)-\left(8+\frac{2}{10}\right) = \left(4-8\right)\color{red}{+}\left(\frac{5}{10}-\frac{2}{10}\right) = -4\color{red}{+}0.3 = \color{red}{3.7}.$$ $\endgroup$
    – Jack D'Aurizio
    Sep 30, 2016 at 15:42
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    $\begingroup$ $4.5 - 8.2 = 4.3 - 8.0 = 0.3 - 4.0 = -(4.0 -0.3) = 3.7$ $\endgroup$
    – Alexis Olson
    Sep 30, 2016 at 15:44
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    $\begingroup$ @JackD'Aurizio That helps a lot. Thank you. One thing- the final result should be -3.7, but again, thank you. $\endgroup$
    – Jim Clay
    Sep 30, 2016 at 15:50
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    $\begingroup$ I disagree that he made a mistake (in arithmetic). He understood what the decimal system is supposed to do, and made a natural move given that understanding, that happens to conflict with standard notation. It's as though the understanding of numbers was strong enough to over-ride the notation, which is good. Of course one can explain the relation to the notation everyone else uses, but working out the problem in his way looks to me like a feature, not a bug. $\endgroup$
    – zyx
    Sep 30, 2016 at 19:52
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    $\begingroup$ Irrelevant side rant: I really don't like the way the problem is stated. Somebody was trying to use math words/variables for their own sake; but "solve for $x$", when $x$ is already isolated, can only lead to misunderstandings about what "solve" means. I'd prefer simply "Compute $4.5-8.2$." $\endgroup$
    – Greg Martin
    Oct 1, 2016 at 19:30

11 Answers 11

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Being the curious kid that he is he asked why the two results weren't the same, and I couldn't give him an answer.

That's because the two results are the same, and he is implicitly using a slightly different and context-dependent notation to express his answer.

The arithmetic is correct, but $-4$ is not a decimal digit in the usual scheme of things.

A correct answer of $(-4).3$ was found, with an intended meaning of $-4 +0.3$. That notation is non-standard, and writing it as $-4.3$ gives the wrong answer when read as a standard decimal.

Although it's clear what an expression like $(-4).3$ should mean here, to represent that result in the standard system with digits 0-9, the minus sign can only apply to all digits in the number at once. The conversion to standard notation is $("-4").3 = -(3.7) = -3.7 $

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    $\begingroup$ If one is looking for some phrase that "corrects the mistake", it is that: in the standard system ... the minus sign can only apply to all digits at once. $\endgroup$
    – zyx
    Sep 30, 2016 at 20:12
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    $\begingroup$ The result may be written as $\bar{4}.3$. This is the bar notation used in tables of logarithms. It is mentioned in Wikipedia article on common logarithm. $\endgroup$
    – Kamil Maciorowski
    Sep 30, 2016 at 20:20
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    $\begingroup$ For integers it works, but with decimals, does $5.\overline{4}$ represent (5 - 0.4), or 5.44444444.....? $\endgroup$
    – zyx
    Sep 30, 2016 at 20:28
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    $\begingroup$ It's all about conventions. In my education (in Poland) I have never encountered any representation like $5.\bar{4}$. It was always $5.(4) = 5.444444…$ (well, if fact $5,(4)$ as we use comma instead of dot – it's yet another convention thing). $\endgroup$
    – Kamil Maciorowski
    Sep 30, 2016 at 20:37
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    $\begingroup$ Certainly if one has different notation for infinite repetition and for negation, the two can be combined in a painless way. Some additional rules are needed to make it compatible with $\bar{z}$ for complex number conjugation. $\endgroup$
    – zyx
    Sep 30, 2016 at 22:11
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He's trying to solve this problem:

$$\begin{equation} \frac{ \begin{array}[b]{r} 4.5 \\ - 8.2 \end{array} }{ } \end{equation}$$

The $.5-.2$ part doesn't cause any heartburn, but when he gets to the $4-8$ part, he'll need to borrow from the tens column. But there's nothing to borrow from!

When performing the subtraction algorithm, the minuend has to be greater than or equal to the subtrahend.

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    $\begingroup$ My attempt to subtract directly ended up with 6.3 - 10, where the 10 represents the unresolved borrowing. It's still an expression with two terms, but if you have a method of calculating "ten's complement", it could be a path to the answer (6.3-10 = -3.7). $\endgroup$
    – AshleyZ
    Sep 30, 2016 at 21:03
  • $\begingroup$ Good emphasis that he is performing an algorithm and wondering why it fails. $\endgroup$
    – PJTraill
    Oct 1, 2016 at 10:49
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It seems the minus in \begin{align*} 4.5-8.2 \end{align*} looks so dominant. We are tempted to think

  • at first we subtract $8$ from $4$ giving $4-8=-4$

  • and then we have to subtract $0.5-0.2$ resulting in $-4-0.3=-4.3$

But the correct way is

  • at first we subtract $8$ from $4$ giving $4-8=-4$

  • and then we have to add $0.5-0.2$ resulting in $-4+0.3=-3.7$

We have to add the values since \begin{align*} 4.5-8.2&=(4\color{blue}{+}0.5)-(8\color{blue}{+}0.2)\\ &=(4-8)\color{blue}{+}(0.5-0.2)\\ &=-4\color {blue}{+}0.3\\ &=-3.7 \end{align*}

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how about draw the number line? That helped me when I was little. When 4.5-8.2, you first -4.5 to get to the 0 point, and you need another part keeping going to reach the total length 8.2. so that will be 8.2-4.5=3.7 away from 0 on the negative side. Hope it will help.

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Performing subtraction one has to do two calculations. At first to determine the sign of the difference by comparing the sizes of the minuend and the subtrahend. Example: the sign of $$3-5$$ is negative as $5$ is greater than $3$. So by now we have $$3-5={-}$$

And know you calculate how much $5$ is greater than $3$, namely $5-3=2$. So finally $$3-5=-2.$$

As for $4.5-8.2$ we have firstly a negative sign and, as $8.2-4.5=3.7$, we arrive in $-3.7$.

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  • $\begingroup$ What was the downvote for? $\endgroup$
    – Michael Hoppe
    Oct 1, 2016 at 7:31
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==== entirely new answer =====

Before we start let's consider an idea I'm going to call "10s complement"

Consider:

6 + ? =10

27 + ?? = 100

284369 + ?????? = 1000000

In other words: xyz....efgh + ???....???? = 1000.....0000

The answer's actually kind of slick.

abcd + (9-a)(9-b)(9-c)(10-d) = 10000

Here's why:

d + (10-d) = 10 of course, so we carry the 1.

c + (9-c) = 9 but we carried the one so ... 10 and we carry the 1.

b + (9-b) = 9 and .... well, you get it ....

...and it goes on till the very end.

So for every n-digit decimal number $abcdefg$ there is a distinct n-digit "complement" $hijklmn$ so that together they add up to 1 with n zeros.

[Now, if you were paying attention, it should be pretty clear that "10 complement" of $abcde$ is the same thing as $(100000 - abcde)$].

........ back to the original subject ......

So consider if we attempted to do what you son did and subtracted a large number directly for a small number.

We'd get

2941

-8472

====

1-2 is 9, borrow the 1. 3 -7 is 6, borrow the 1. 8 minus 4 is 4. 2 minus 8 is 4 borrow the 1 ... um, wait, what... where do we borrow the one from?...

2941

-8472

[-10,000] + 4469

with a final -1 borrowed from ... the universe. This means we are "in debt" of 1 in the 10,000s column. We "borrowed" 10,000 from a place we just don't have.

So we will have a negative number. It is the negative number that 4469 + ????=10,000.

It's the "10s complement" I began the post with!

So the answer is, you can subtract directly. But you'll need to borrow a power of 10 (1 with some zeros) to get the negative of the "10s complement"

So.... 2941 - 8472 = 4469 - 10000 = -(9-4)(9-4)(9-6)(10-9) = -5531 = -(8472-2941)

====old answer ====

Well, the logic that they can't be the same because:

1) small - big = -(big - small)

so if 2) subtracting directly were okay that would mean

3) subtracting (big - small) indirectly would have to also be true.

e.g. If we could subtract 4.5 - 8.2 by subtracting 5 -2 directly then we could, when facing 8.2 - 4.5, CHOOSE if we want to subtract the 5 from the 2 or the 2 from the 5, and obviously we know that those aren't the same...

but why do we know that?

Consider this alternative way of doing subtraction without borrowing along the way (instead we borrow at the very end):

816

-542

===

3(-3)4 =

300 - 30 + 4 =

(300 - 100) + (100-30) + 4 =

= 200 + 70 + 4

274.

Then if we subtracted (small - big) we'd get.

542

-816

===

(-3)3(-4)

=-300 + 30 -4

=[-1000] + (1000-300) + (30 - 10) + (10-4)

= [-1000] + 700 + 20 + 6 = [-1000] + 726

And we have big honking 1000 borrowed from ... the universe. This will never resolve.

....

Now lets step back from this and consider all sorts of problems like:

6 + ? = 10

26 + ?? = 100

2346839684561 + ????????????? = 10000000000000

All these types of problems have a sort of "taking the opposite" approach.

You figure what added to the first makes 10. Then with carrying you from what added to the rest make 10

3236 + ???? = 10000

???? = (9-3)(9-2)(9-3)(10 - 6) = 6764. 6764 and 3236 are "opposites" in that pairwise the add so 10000 or some power of 10.

So back subtraction.

The "opposite" of 726 is 274 . So the answer is -274.

So you can subtract directly. But in the end you'll have to take the "opposite"

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    $\begingroup$ This may be interesting, but I was put off by the length and the formatting — perhaps an introductory summary would help. $\endgroup$
    – PJTraill
    Oct 1, 2016 at 10:53
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He subtracted the decimal part twice, once to get the 0.3 and then again to go from -4 to -4.3 -- he should have added the 0.3 to -4 to get the final answer.

You can subtract wherever you want, but you must only do it once. Otherwise, you're adding rather than subtracting.

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    $\begingroup$ He effectively subtracted twice, but psychologically he had a different problem (or he was just blindly performing the algorithm). $\endgroup$
    – PJTraill
    Oct 1, 2016 at 10:55
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The standard subtraction algorithm only works when the larger-magnitude number is on top (in this case, the $8.2$). The most straightforward calculation is to note that the result here will be obviously negative (because the number being subtracted is of larger magnitude), and then perform $8.2 - 4.5$ for the magnitude of the result.

Failing to pay attention to which magnitude is larger results in exactly this kind of subtraction error.

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Expanding on @John's answer https://math.stackexchange.com/a/1948174/169560. You can borrow from "nothing" you just need to remember to give it back:

\begin{equation} \frac{ \begin{array}[b]{r} 4.5 \\ - 8.2 \end{array} }{}= \frac{ \begin{array}[b]{r} \ ^{-1}14.5 \\ - 8.2 \end{array} }{}= \frac{ \begin{array}[b]{r} \ ^{-1}14.5 \\ - 8.2 \end{array} }{-10+6.3}=-3.7 \end{equation}

This way, if the minuend is smaller than the subtrahend, you will end up with something like $-100 \ldots 000 + YY \ldots YY.yy \ldots yy$ where $x$, $Y$ and $y$ stand for any digits, and the number of zeroes is the same as the number of $Y$s. It should be easy to do this last step, as you just need to pad to a potence of 10 - and you train questions like $100 - 22 = ?$ (which some people answer wrongly as $88$) as a bonus.

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Well, too bad nobody uses tape decks these days any more since this is a case of "tape deck counter arithmetic" (which is, as mentioned in other answers, 10s complement arithmetic). 999, 998, 997 ... are dual representations of -1, -2, -3. Your son arrived at 3 for the last digit which is the dual of -7. And in tape counterese, his result was 996.3 which is the dual of -3.7: if you forward your tape from a counter setting of 996.3 by 3.7, it will be zero again.

Whether the last digit is really 3 or 7 is something which you can only decide once you actually know whether the result as a whole should be positive or negative, and you don't know that yet when starting from the right.

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I think that John above was onto something: "There was nothing to borrow from". The real problem is that the subtraction was never completed. If it were, the result would be an infinitely long string of leading nines.

In early calculating machines, even soroban and addiator, subtraction was handled by tens complement addition.

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