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 ?

Easy Math Editor

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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Comments

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 - 8 years ago

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.

Sreejato Bhattacharya - 8 years 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.

Nishant Sharma - 8 years 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.

Sreejato Bhattacharya - 8 years ago

@Sreejato Bhattacharya – Great answer.

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 $n$ .

Calvin Lin Staff - 8 years ago

@Calvin Lin – @ Calvin

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

Nishant Sharma - 8 years ago

@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 - 8 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$