A doubt on division of numbers

This might seem very simple to some but I got confused thinking over it. It goes like this:

Is every rational number divisible by every irrational number ? Why(then which of them are divisible)/ why not ? More specifically what are the criteria for the aforesaid division to be successful ?

Does the vice-versa also hold true ?

Note by Nishant Sharma
6 years, 4 months ago

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4 votes

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When we usually talk about two numbers being divisible, we talk about the natural numbers, or sometimes even the integers. So having rational numbers or irrational numbers divide the other evenly doesn't make sense in all cases.

Bob Krueger - 6 years, 4 months ago

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We know that a a divides b b if and only if there exists some integer k k such that b=ka b= ka . Now let p p be a rational number and let q q be an irrational number. If q q divides p p , we have p=kq p= kq where k k is some integer. This implies q=pk q= \frac{p}{k} . Now we know that every integer is a rational number, so k k is also a rational number. But when a rational number is divided by a non zero rational number, the result is also a rational number. Hence q q is a rational number, a contradiction. But note that we can derive q=pk q= \frac{p}{k} only when k0 k \neq 0 . If k=0 k= 0 , we have p=0 p= 0 . Hence no irrational number can divide any rational number other than 0 0 .

Similarly, assume any irrational number is divisible by a rational number. Then we have q=pk q= pk (notations same as above). But the product of two rationals is always a rational, thus we have a contradiction. Hence no irrational number is divisible by a rational number.

This is what I think. Please correct me if I am wrong.

Sreejato Bhattacharya - 6 years, 4 months ago

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Yeah that was ok. I spoke of this only because I encountered it while solving a problem. The problem goes like this:

Let aa, bb, cc be positive integers such that a divides b2b^{2}, bb divides c2c^{2} and cc divides a2a^{2}. Prove that aabbcc divides (a+b+c)7(a + b + c)^{7}.

Using the b2b^{2} = kaka, c2c^{2} = mbmb and a2a^{2} = ncnc(\big(where kk, mm and nn are some positive integers)\big), I arrived at showing kmnkmn divides kmnkmn (\Big((\big(nnm3m^{3})\big)^17\frac{1}{7} + (\big(kkn3n^{3})\big)^17\frac{1}{7} + (\big(mmk3k^{3})\big)^17\frac{1}{7})\Big)^7. I am not able to show that the factor inside the big braces is an integer. Please help.

Nishant Sharma - 6 years, 4 months ago

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From the multinomial theorem, we know that every term in the expansion of (a+b+c)7 (a+b+c)^7 is of the form mapbqcr ma^pb^qc^r , where m m is an integer and p p , q q , and r r are non-negative integers such that p+q+r=7 p+q+r= 7 and atleast one of p p , q q , or r r is positive. Now note that when all of p p , q q , and r r are positive, abcabc divides mapbqcr ma^pb^qc^r . Now let any one of p p , q q , or r r be zero. By symmetry assume p=0 p= 0 . Then we have q+r=7 q+r= 7 . By symmetry assume q>r q> r (note that since 7 7 is an odd integer pq p \neq q ). Since both q q and r r are positive integers, minimum possible value for q q is 4 4 , while its maximum possible value is 6 6 . If q=4 q= 4 , r=3 r= 3 , and the term is mb4c3=mb2b2c3 mb^4c^3= mb^2*b^2c^3 . Since a a divides b2 b^2 , we get that this term is divisible by abc abc . Similarly we can prove that abc abc divides the term when q=5 q= 5 and when q=6 q= 6 . Now consider the case when two of p p , q q , or r r are 0 0 . By symmetry let p p be non zero. Then we must have p=7 p= 7 . Then the term is ma7=ma4a2a ma^7 = ma^4*a^2*a . Now c c divides a2 a^2 , so c2 c^2 divides a4 a^4 . But b b divides c2 c^2 , so b b divides a4 a^4 . Thus abc abc divides ma7 ma^7 . So we have proved that abc abc divides all the terms in the expansion of (a+b+c)7 (a+b+c)^7 (note that all of p p , q q , and r r cannot be 0 0 ) . Since (a+b+c)7 (a+b+c)^7 is actually the sum of all such terms, abc abc divides (a+b+c)7 (a+b+c)^7 .

Note:- Note that the multinomial coefficient (i.e m m ) must always be an integer. According to the link provided, the multinomial coefficient is n!n1!n2!...nk! \frac{n!}{n_1!n_2!...n_k!} , where n=n1+n2+...+nk n= n_1+ n_2 + ... + n_k . This is equal to the number of permutations of n1 n_1 identical objects, n2 n_2 identical objects, ..., nk n_k identical objects put together. This has to be an integer obviously.

Sreejato Bhattacharya - 6 years, 4 months ago

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@Sreejato Bhattacharya Great answer.

Another approach (which is almost equivalent to yours) that doesn't use the cases, is to argue that since ab2c4a8 a\mid b^2 \mid c^4 \mid a^8, hence the set of primes which divide a,b,a, b, and cc are the same. For a given prime kk, let kak_a, kbk_b and kck_c be the number of times that kk divides into a,ba, b and cc respectively. Then, we know that the number of times kk divides into abcabc is ka+kb+kc k_a + k_b + k_c . We also know that ka2kb4kc8ka k_a \leq 2 k_b \leq 4 k_c \leq 8 k_a . Hence, for any integer triples such that x+y+z=7 x+y+z = 7 , we have xka+ykb+zkc7min(ka,kb,kc)ka+kb+kc x k_a + y k_b + z k_c \geq 7 \min(k_a , k_b, k_c) \geq k_a + k_b + k_c . This shows that for each prime pp, the power of pp which divides abc abc is smaller than (a+b+c)7(a+b+c)^7, Hence, abcabc divides (a+b+c)7(a+b+c)^7 .

Notation: Given a prime pp and an integer nn, we say that pp divides into nn kk times if pk p^k divides nn but pk+1 p^{k+1} doesn't divide nn. As such, pkp^k is the highest power of pp that divides nn.

Calvin Lin Staff - 6 years, 4 months ago

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@Calvin Lin @ Calvin

Yup..I couldn't follow your last few statements.

Nishant Sharma - 6 years, 4 months ago

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@Sreejato Bhattacharya @Sreejato

Ohh..... I missed that fact(courtesy my negligence towards combinatorics). Now with your help I have been able to solve it. TY.......

Nishant Sharma - 6 years, 4 months ago

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