When you divided a rational number by an irrational number...?

Question

is the result always irrational?

Answer 1

Not necessarily.

0/pi = 0, which is a rational number.

Let's try a proof with nonzero rational numbers.

Claim: Assume m is rational and n is irrational. Then
m/n is irrational for non-zero m.

Proof: (by contradiction)

Assume m/n is rational. Then m/n can be expressed as the quotient of two integers a and b; that is

m/n = a/b, for b nonzero. Cross multiply,

mb = na

Since m is nonzero, it follows that a must be nonzero, so we may divide both sides by a.

mb/a = n

m is a rational number, so m can be expressed as c/d, for c and d integers.

(c/d)b/a = n

(bc/d)/a = n
(bc)/(ad) = n

The integers are closed under multiplication (the product of two integers is an integer), which means we have a quotient of integers. This implies n is rational.

But n is irrational, so this is a contradiction.

Therefore, your statement holds true if the rational number is non-zero.

Answer 2

Yes. Call the irrational number x, the rational number a/b. Then, if a/bx = c/d for some rational number c/d, it would follow that x = ad/bc. But, that would make x a rational number, a contradiction.

OK, as Puggy alertly notes, the exception would be if the rational number is zero. :)

Answer 3

If a rational number, x, is not zero it can be written as p/q where p,q are integers.

be y another number such that x/y is rational. In this case there exist to integers m,n such that x/y = m/n

x/y = (p/q)/y = m/n

p/q = my/n // multiply by n/m

y=pn/qm

pn and qm are both integers, thus y is rational.

So, the answer to your question is yes.

Answer 4

yes