Why is there no “remainder” in multiplication (math.stackexchange.com) 1 points by patrickg 13y ago ↗ HN
[–] lutusp 13y ago ↗ Of course there are remainders in multiplication. Please don't tell your son anything different.Example: 5 1/3 x 5 = 26 2/3Another example 1/7 * 23 = 23/7There are any number of such examples.
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[ 3.0 ms ] story [ 12.2 ms ] threadExample: 5 1/3 x 5 = 26 2/3
Another example 1/7 * 23 = 23/7
There are any number of such examples.