Can anyone explain to me "subtraction by addition" in a visual way?
The steps say:
- Take the "complement" of the number we are subtracting
- Add it to to the number we are subtracting from
- Discard the extra "1" on the left
Example: $9 - 7$
- $7$'s complement is $3$
- Therefore $9 + 3 = 12$
- We discard the $1$ on the left and the answer is $2$
I don't understand how it actually works. Please help me to understand it.
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$\begingroup$$$9-7=9-(10-3)=9+3-10=12-10=2$$
$\endgroup$ 3 $\begingroup$As you have asked for a "visual" explanation:
The "complement" of $7$ is the number of steps it needs to reach $10$ (that is $3$). But on the way to reach $10$, it reaches $9$ first. Thus, of the $3$ steps, $2$ steps are needed to reach $9$, and the remaining $1$ step takes $9$ to $10$. So, when you add $3$ to $9$, the first thing that happens is that $9$ itself reaches $10$ in that $1$ step, and then it goes above $10$ with the remaining $2$ steps. When you remove $10$ from this answer (by discarding the "extra $1$ on the left"), you get the $2$ steps which $7$ actually takes to reach $9$.
It was amusing to explain this, but I agree with Thomas in finding this method unnecessary and crippled.
$\endgroup$ 4 $\begingroup$$$9-7 = 10-8 = 11-9 = 12 - 10$$
$\endgroup$ $\begingroup$The method you are using is what we call Complement's Method or Method of Complement.
This method is implemented most commonly in digital computer to perform binary arithmetic.
Processes in complements method (of decimal number)
- negative number is represented in complemented form of either 9 or 10.
For example,if a number is -15,
9's complement of 15 is 84, because 9-1=8 and 9-5=4
10's complement of 15 is 84 + 1 i.e. 85.
Note: The nines' complement of a decimal digit is the number that must be added to it to produce 9; the complement of 3 is 6, the complement of 7 is 2, and so on. 10's complement is 9's complement plus 1.
- positive number is left as it is
- addition is performed between the positive number and complemented form of negative number
- if 9's complement is performed, then 1 is added to final answer
if 10's complement is performed, the sum is left as it is. - if extra digit appears in the final answer, it is omitted.
For example, for 50 - 5
9's complement:
- 50 - 5 → 50 + 94 (5 is taken as 05 as highest number of digit is 2)
- 50 + 94 = 144
- 144 + 1 = 145
- 145 → 45
10's complement:
- 50 - 5 → 50 + 94 + 1 → 50 + 95
- 50 + 95 = 145
- 145 → 45
I've been learning how to use the Japanese abacus for some years. So I'm used to calculate with complements all the time both when I add numbers or when I subsctract.
But this example "9 - 7" doesn't look particularly good to me. Because if I've got 9 beads on a row I don't need any trick to remove 7 beads: I just put my fingers on them and when I remove 7 beads, I've left with 2, end of story.
But if the exemple were "10 -7" and I have 1 bead on the tens and no beads on the one's, then I can't remove 7 beads from the one's row. So here it makes totally sense using complements.
I will borrow one from the tens row (by removing it) and then add 3 on the ones row, because 3 is the complement of 7 .
if I have the number 10 and I remove the one on the tenths row and I add 3 on the units row, the final result is = 3
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