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So, say you have a really huge whole number like $5^{2000}$ and an irrational number like $\sqrt(5)$.

If you were two multiply the two would you get an even or odd number after rounding to the nearest whole number? $$5^{2000} * \sqrt(5)$$

I realize I could input this into a calculator and have it tell me, but is there a general why to know whether the answer would be even or odd?

I'm thinking it has something to the number of digits in the whole number, but I can't think of a way to guaranteed method(I've tried looking this up ): ). How many digits of the irrational number would I need?

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closed as unclear what you're asking by Peter, José Carlos Santos, Robert Z, stressed out, Shailesh Feb 17 at 0:04

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    $\begingroup$ You get an irrational number, which is neither even nor odd. $\endgroup$ – Torsten Schoeneberg Feb 15 at 0:36
  • $\begingroup$ We generally only talk about parity of numbers (even vs odd) when talking about integers or whole numbers. E.g. $5$ is odd, $20$ is even, $3^{1000}$ is odd, etc... We do not talk about fractions being even or odd. $\frac{1}{2}$ is neither even nor odd, it is not an integer and so does not receive either label. Similarly $\sqrt{5}$, being an irrational number, is not an integer and so also does not get the status of being called either "even" or "odd." Irrational times integer is always irrational, even if that integer is huge. $\endgroup$ – JMoravitz Feb 15 at 1:12
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    $\begingroup$ As written, it is hard to understand exactly what you are thinking with this question. It sounds like you might be implying that if you multiply an irrational number by a large enough integer that the result will be an integer and so can be called even or odd which is of course utter nonsense. If that were possible then the number wouldn't have been irrational in the first place. $\endgroup$ – JMoravitz Feb 15 at 1:14
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    $\begingroup$ Sorry! I meant if I were to round the product to the nearest whole number afterwards. $\endgroup$ – Wilson Guo Feb 15 at 1:58
  • $\begingroup$ I would like to see you input $5^{2000}\sqrt5$ into a calculator. $\endgroup$ – Gerry Myerson Feb 15 at 3:19
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I presume you are asking about the parity of the whole number you get by rounding or truncating $5^{2000}\sqrt 5$. The issues are the same whether you round or truncate, though the number you get may be different and the answer may be different.

$5^{2000}$ is a number of $\lceil 2000 \log_{10}5\rceil=1398$ digits. In fact, it is rather close to $10^{1398}$ If we want to know $5^{2000}\sqrt 5$ to within a possible error of $\pm 0.5$ we need to know $\sqrt 5$ to within $\pm 0.5\cdot 10^{-1398}$, which means you need $1399$ digits of $\sqrt 5$ past the decimal point.

There are perverse cases where the error you can tolerate is much smaller than $\pm 0.5$. It happens when the exact product is very close to the breakpoint between two rounded values. If we are truncating and the correct answer might be $2.999999$ and might be $3.000001$ a difference of only $0.000002$ is enough to change the result. The nature of truncating or rounding makes it impossible to avoid this.

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