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# 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
3 years, 9 months ago

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

Similarly, assume any irrational number is divisible by a rational number. Then we have $$q= pk$$ (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. · 3 years, 9 months ago

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

Let $$a$$, $$b$$, $$c$$ be positive integers such that a divides $$b^{2}$$, $$b$$ divides $$c^{2}$$ and $$c$$ divides $$a^{2}$$. Prove that $$a$$$$b$$$$c$$ divides $$(a + b + c)^{7}$$.

Using the $$b^{2}$$ = $$ka$$, $$c^{2}$$ = $$mb$$ and $$a^{2}$$ = $$nc$$$$\big($$where $$k$$, $$m$$ and $$n$$ are some positive integers$$\big)$$, I arrived at showing $$kmn$$ divides $$kmn$$ $$\Big($$$$\big($$$$n$$$$m^{3}$$$$\big)$$^$$\frac{1}{7}$$ + $$\big($$$$k$$$$n^{3}$$$$\big)$$^$$\frac{1}{7}$$ + $$\big($$$$m$$$$k^{3}$$$$\big)$$^$$\frac{1}{7}$$$$\Big)$$^7. I am not able to show that the factor inside the big braces is an integer. Please help. · 3 years, 9 months ago

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

Note:- Note that the multinomial coefficient (i.e $$m$$) must always be an integer. According to the link provided, the multinomial coefficient is $$\frac{n!}{n_1!n_2!...n_k!}$$, where $$n= n_1+ n_2 + ... + n_k$$. This is equal to the number of permutations of $$n_1$$ identical objects, $$n_2$$ identical objects, ..., $$n_k$$ identical objects put together. This has to be an integer obviously. · 3 years, 9 months ago

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

Notation: Given a prime $$p$$ and an integer $$n$$, we say that $$p$$ divides into $$n$$ $$k$$ times if $$p^k$$ divides $$n$$ but $$p^{k+1}$$ doesn't divide $$n$$. As such, $$p^k$$ is the highest power of $$p$$ that divides $$n$$. Staff · 3 years, 9 months ago

@ Calvin

Yup..I couldn't follow your last few statements. · 3 years, 9 months ago

@Sreejato

Ohh..... I missed that fact(courtesy my negligence towards combinatorics). Now with your help I have been able to solve it. TY....... · 3 years, 9 months ago