I apologize in advance as this post will be clumsily written and likely require only one response—likely from an 8th grader.

I know that given a decimal ‘x’ with n digits, and a decimal ‘y’ with m digits, the product (x*y) will have at most m + n digits.

What I’m concerned with is numbers after the decimal point.

I think I’m right, I just can’t convince myself one way or the other.

Given decimals ‘x.y’ and ‘z.w,’ if ‘y’ has n digits, and ‘w’ has m digits, will the product ([x.y]*[z.w]) have at most m + n digits past the decimal place?

In this case, I’m actually concerned with the number of digits after the decimal. It may be the case that I’m actually talking about significant digits (or that I can answer the underlying question with same) but what I’m unsure of is something like this (again, apologies for poor math notation):

If x*y will have at most n + m digits, then .x * .y will have at most n + m digits past the decimal. (All assuming the points made in the OP.)

But I get lost (or unable to convince myself) after this:

Given decimals with however many total digits, AND m and n digits after the decimal place (respectively) will the product necessarily have, AT MOST, m + n digits after the decimal place?

[ul]
[li]x has a non-zero digits to the left of the decimal place, and b to the right. (i.e. x = <a digits>.<b digits>).[/li][li]y has c digits to the left of the decimal place, and d to the right.[/li][/ul]

Then the numbers x * 10^b has a+b non-zero digits to the left of the decimal, and none to the right. Similarly, y * 10^d has c+d non-zero digits to the left of the decimal, and none to the right.

Since both these numbers are integers, by your previous theorem, the product of these numbers has, x * y * 10^(b+d) at most a+b+c+d non-zero digits to the left of the decimal (and none on the right).

Dividing this by 10^(b+d) to get x*y shifts the whole number b+d places to the right. Since x * y * 10^(b+d) is an integer, this means that x * y has at most b+d non-zero digits to the right of the decimal. QED.

No, I think (s)he’s talking about something more basic than that. Significant digits are only involved when you’re dealing with numbers that are approximations of something.

I’m pretty sure the answer to the OP’s question is “yes,” and for essentially the same reason as for the whole number case.

If you multiply this by another number with n digits after the decimal point, it’ll have a digit that represents 10[sup]–n[/sup]; and multiplying these together will give you a term with 10[sup]–3–n[/sup] plus other terms with higher powers of 10; so that last digit will be (at most) 3+n places after the decimal point.

Think about the procedure for multiplying decimals. The number of the decimals places in the product is equal to the sum of the decimal places in the factors.

Mathematical (pseudo)proof.
let x.y = a x 10^-n and x.w = b x 10^-m such that a and b is an integer not divisible by 10. This is equivalent to what the OP proposes.

x.y times z.w = a times b times (10)^(-m + -n)
a times b is an integer and multiplied by 10^-(m+n) means that there are m+n decimal places or less if a times b is a multiple of 10^t if t is a natural number. Certainly never more than m+n decimal places.

The procedure I described is correct but you can lose trailing zeros as well. The point being there is no way to end up with more decimal places. Notice I address your point in the second part of my post.

True, except that the lowest order digits might be zeros. For example: .2 * .5 = .10

Remembering back to the fixed-point computer arithmetic I did for auto engine control computers, we had to track the number of significant digits on both sides of the radix point very carefully, but we didn’t worry that low-order ones might be zeros. You only worry about that when writing the number for human eyes. In that case, the simple formula doesn’t quite work.

Bingo. It sheds light on the issue here as a rhetorical question, but isn’t answerable.

Right; that’s why I, and the OP, were careful to say “at most” when talking about how many digits there would be.

To take an extreme example, it’s possible to multiply two numbers that have infinitely many digits after the decimal point and get a product that has 0 digits after the decimal point.sqrt(2) * sqrt(2) = 2, for example