Why is there no “remainder” in multiplication | Hacker News

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		Why is there no “remainder” in multiplication (math.stackexchange.com)
		1 point by patrickg on Oct 1, 2013 \| hide \| past \| favorite \| 1 comment

lutusp on Oct 1, 2013 [–]

Of course there are remainders in multiplication. Please don't tell your son anything different.

Example: 5 1/3 x 5 = 26 2/3

Another example 1/7 * 23 = 23/7

There are any number of such examples.

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