why infinite carrying does not occur

Mathematics Asked by Ken.Wong on December 12, 2020

This is probably a trivial question in arithmetic. Given a number $$x=0.a_{1}a_{2}…$$ which never terminate. If we do a multiplication with an integer $$N$$, in real life of course we truncate the the $$x$$ and doing it using computer or something. But how do we know that the part being truncated doesn’t really matter at all? i.e. When $$N$$ times the truncated part, since there are infinite digit, and if $$N$$ is large, there will be carrying occur at each digit and hence an infinite carrying occur, which affect the result with infinite carrying. However this does not occur. Because it seems to me that in daily life, when we truncate our expression, the truncation only affect a few digit at all. It seems we are safe to truncate infinite digit without worrying it affect the whole expression. So why does this not occur?

Truncating a decimal number with infinitely many digits may seem risky, but if it does done with care, then few problems should arise. For instance, $$piapprox3.14$$. Let's imagine I didn't know the next digit of $$pi$$. Just from this truncation, I know that

$$3.14

From this I know that $$pi$$, multiplied by $$10$$, must be between $$31.4$$ and $$31.5$$:

$$31.4<10pi<31.5$$

Therefore I can say with confidence that I know the value of $$10pi$$ to the nearest whole number. The fact that there are infinitely many digits after $$3.14...$$ is actually less of an issue than you might think. One intuitive way to think about this is that each digit of a number is $$10$$ times less important than the previous digit. The potential impact that a new digit could have decays exponentially. The next digit after $$3.14$$ can only make the number bigger at most by $$9/1000$$. By the time you get to $$3.141592$$, you are already within one-millionth of the true value of $$pi$$. Is it any wonder, then, that even NASA only uses the measly $$3.141592653589793$$ for its astronomical calculations?

Therefore, multiplying a truncated decimal by another number shouldn't pose too much of an issue, unless you are multiplying by a very large number. If, for instance, you were multiplying $$pi$$ by $$100 000 000$$, then it would be sensible to work out the accuracy of the approximation in a similar fashion to the way I did above.

Answered by Joe on December 12, 2020

You can indeed wind up with infinitely many carries! Consider for example $$0.555555...times 2,$$ in which each digit past the decimal point winds up producing a carry. However, this doesn't affect our ability to make sense of this product nonetheless (and indeed the above is just $$1.11111...$$ - or, switching from decimals to fractions, $${5over 9}times 2={10over 9}$$).

The key point is this: even though infinitely many carries are happening "overall," each specific digit of the answer is only affected by finitely many carries. In this case, for example, each digit "absorbs" one carry. It's a good exercise to whip up an upper bound on how many digits we have to look at to determine a given digit of a product $$0.a_1a_2a_3...times k$$ in terms of $$k$$.

• HINT: first assume $$k$$ is a natural number, and consider the number of digits in $$k$$ itself.

• COMMENT: note that I said "whip up an upper bound" - I didn't say that you should also prove that your upper bound works. This is because this proof is quit tedious if you don't come at it the right way, but that right way is fairly abstract: we need to go to the formal definition of infinite decimal expansions, and I think it winds up being too hard for an exercise. After you familiarize yourself with infinite series, and in particular become comfortable with the idea that "$$0.a_1a_2a_3...$$" is shorthand for "$$sum_{i=1}^infty a_itimes 10^{-i}$$," you'll have the necessary tools: the argument will amount to just bounding the appropriate partial sums.

Answered by Noah Schweber on December 12, 2020

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