4

Possibly a question for the math or educational crowd, but...

Given

          4
       x 12
     ------

Is there a specific term for the "top" number? Other than first or top of course.

EDIT: I would read the above as "four times 12"

  • 1
    The top one is the multiplier, the bottom the multiplicand. – Mitch Jan 11 '15 at 20:50
  • I always took it the other way around. Not that it matters. – Hot Licks Jan 11 '15 at 21:22
  • 4
    @Mitch When I was at school, aged about 12, the maths teacher explained that one was the multiplier and the other the multiplicand. In the nearly sixty years since then I have never benefitted in any way from knowing that, notwithstanding a career in accountancy. So it is with a sense of nostalgia that I read your comment. – WS2 Jan 12 '15 at 0:41
  • 1
    I'm remembering that when I was working at RCA in the early 70s, writing "horizontal microcode" to implement the arithmetic operations of a S/360-style machine architecture, we simply called both operands "multipliers". There was absolutely no reason to distinguish one from the other. – Hot Licks Jan 12 '15 at 3:10
  • 1
    I have a degree in mathematics, and I've never heard of one factor being called the multiplier and the other the multiplicand. It seems like pointless terminology to me since the order of multiplication doesn't matter here. Worse, it is ambiguous since both factors could be called multipliers or multiplicands. – dangph Jan 12 '15 at 8:11
12

Depending on how one interprets the vertical notation (is that twelve times four or four times twelve?) it can be either the multiplier or the multiplicand; and the solution is termed the product.

  • You beat me :-( – Matt Gutting Jan 11 '15 at 20:39
  • 1
    +1, but I think we can resolve that "depending on…" down to a single answer. – Jon Hanna Jan 11 '15 at 21:08
4

I would call it the first factor. In this particular example, we have two factors (4 and 12). Their product being 48.

2

It's the multiplier.

The vertical format can only be considered equivalent to 4 × 12 because while multiplication is commutative and hence 4 × 12 is the same as 12 × 4, subtraction is not and when writing the subtraction 12 - 4 in vertical form we put the 4 at the bottom.

Hence the top number in this format must be considered the multiplier, and the second must be considered the multiplicand.

  • 2
    I seem to remember, though, that in multiplying multi-digit numbers with pencil and paper the practice I learned in school effectively treated the lower number as multiplier. If the two factors were 12 and 349, regardless of the order in which they were given, we would put the 12 on the bottom, and then in effect ask ourselves first what was 2 times 349 and then what was 10 times 349, then add those results together. – Brian Donovan Jan 11 '15 at 21:16
  • 1
    I believe this answer is exactly backwards. The bottom number in subtraction is the subtrahend, that which is subtracted (from the minuend). In multiplication, the multiplicand is the number which is multiplied (by the multiplier). In multiplication, the figure which is multiplied is the one on top. Commutativity makes no difference to the names: change the order of the numbers, and they change their names -- names are based on position. A comparison with subtraction is only valid to that extent. – Andrew Leach Jan 12 '15 at 7:39
  • @AndrewLeach of all the dictionary definitions I can find, only wiktionary and the OED are explicit about position, and they both agree that the first one is the multipler. "A comparison with subtraction is only valid to that extent" That's why I made a comparison with subtraction to that extent. – Jon Hanna Jan 12 '15 at 9:51
  • See also this. The answer there quotes a supposedly reputable source saying 5 + 5 + 5 ⇒ 3 × 5. I would say that should be 5 × 3 (because you start with a five and write three of them); I agree with that maths tutor and his analysis of the "vertical" sum. – Andrew Leach Jan 12 '15 at 10:01
  • @AndrewLeach eh, the question is asking why that same reputable source agrees with me… – Jon Hanna Jan 12 '15 at 10:07

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