I wanted to know if two non-int numbers (non-zeroes) when divided with each other can give an integer or not.I believe that's a NO. However I know they can only yield an integer $1$ provided both are the same (i.e) $\frac{2.5}{2.5}$. Am I correct. Just wanted to be 100% sure. How about when they are multiplied ?
Can dividing two rational numbers yield an integer?
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arithmetic
rational-numbers
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0hint: an integer is a rational number – 2012-07-27
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3You can certainly divide a rational by a rational and get an integer. Example: $\frac{2/7}{1/14}=4$. If we have two numbers, rational or not, such that $\frac{r}{s}$ and $\frac{s}{r}$ are **both** integers, then $r=\pm s$. – 2012-07-27
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1That is not what that wikipedia page says. Remember that integers are also rational numbers. – 2012-07-27
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0I actually meant to ask about non-integers - Edited my question. – 2012-07-27
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1The Wikipedia article also doesn't say "no", what it actually says is that dividing one rational by another never produces something that isn't rational. So you can't get irrationals out of rational division. (Don't forget Raymond Manzoni's comment above at this point). – 2012-07-27
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0Can breeding two dogs yield a rottweiler? – 2012-07-27
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0unless both of them are a rottweiler. NO – 2012-07-27
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0@AndréNicolas I actually wanted to ask about non-ints like $\frac{2.6}{5.5}$ – 2012-07-27
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1$\frac{2.4}{1.2}=2$, so the answer to your question is yes. – 2012-07-27
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0If $r$ and $s$ are rationals, $\frac{r}{s}$ is sometimes an integer, and sometimes not. – 2012-07-27
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1@MistyD : Its not the point of integers or rationals. Its the point of G.C.D . If the numerator and denominator have a G.C.D of 1, they wont yield integers and leave you with some decimal part – 2012-07-27
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0Thanks @Iyengar that clears it up also Joel Reyes example helped – 2012-07-27
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0@MistyD : That don't work always. I wanted to edit further, but suddenly my internet was interrupted. Anyway you will not get an integer always when G.C.D is not 1. Suppose $\dfrac{42}{56}=\dfrac{3}{4} \notin \mathbb{Z}$ . But $\rm{G.C.D(42,56)}=14$. So here are few steps you need to work out. 1) See that numerator is always greater or equal to denominator. 2) Reduce the fraction to the least form by applying recursive cancellations. A pseudo code can be as follows. Contd.. – 2012-07-27
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0@MistyD : .. Contd . 1) Start with $ K= \dfrac{a}{b}$. 2)$ M= \rm{G.C.D}(a,b) $3 ) $K=\dfrac{\dfrac{a}{M}}{\dfrac{b}{M}}$. Recursively repeat until it yields an integer 4) If it don't yield an integer take the new numerator and new denominator, if their G.C.D is 1 , then its division will not be an integer. Or else it would be an integer. – 2012-07-27
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0@MistyD : Simply to state $a=n*b$ then it means $ b | a $ for some $n \in \mathbb{z}$ – 2012-07-27
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0What's three-fifths divided by three-tenths? – 2012-07-27
2 Answers
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Of course you can:
2.2/1.1 = 2
2.2 and 1.1 are both obviously rational non-integers, and 2 is an integer. Is this what you wanted?
You can even divide two irrational numbers and get an integer:
(2 * Pi) / Pi = 2
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Let $n$ be an integer and $p:=\frac{n}{n+1},q:=\frac{1}{n+1}$ then $p/q=n$. Note that both $p$ and $q$ are non-integers (as $\gcd (n,n+1)=1$).
Also if $r:=\frac{n^2}{n+1},s:=\frac{n+1}{n}$ then $rs=n$. Clearly both $r$ and $s$ are non-integers, because $\gcd (n^2,n+1)=\gcd (n,n+1)=1.$
So for every integer $n$ you can find non-integral rationals whose quotient is $n$ and non-integral rationals whose product is $n$.