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For simplicity I will assume you're working in base x, for any integer x greater than 1, although the argument extends to integers greater than 1 in absolute value (note that in base -1,1 all decimal numbers are in fact integers and that in base 0 decimals are not very well defined). In base x, x can of course be conveniently denoted as 10, so in the remainder of this answer I will work in base x. It is sufficient to show that there exists a decimal number that is not an integer so take 0.1 or 10^-1. This number has the property that 10*0.1 = 1, it is the multiplicative inverse of 10. I will now prove by induction that no positive integer has this property.

Base case: 1*10 = 10 which is greater than 1 by assumption.

Suppose n*10 is greater than 1, then (n+1)*10 = n*10+1*10 = n*10 + 10 which is still greater than 1.

So we now know that n*10 is always greater than 1 for any n greater than 0, from which it can be deduced that for these n, n*10 is also unequal to -1. Therefore, for no integer n unequal to zero can n*10 be 1. Now assume n=0, then n*10 = 0*10 = 0 which is not equal to 1 either.

Thus, no integer n has the property n*10=1, whereas the decimal number 0.1 does. So 0.1 is not an integer and therefore the decimal numbers are not integers.

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13y ago

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Q: Why are decimal numbers not integers?
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