Question

Ryiaz claims it is impossible to prove that a rational number multiplied by an irrational number gives an irrational product. How can you prove this

Answer 1

To prove that the product of a non-zero rational number and an irrational number is irrational, we can approach the problem as follows:

Let \( r \) be a rational number and \( x \) be an irrational number. We want to show that the product \( r \times x \) is irrational.

Since \( r \) is a rational number, we can express it in the form:
\[
r = \frac{a}{b}
\]
where \( a \) and \( b \) are integers and \( b \neq 0 \).

Now, we will consider the product:
\[
r \times x = \frac{a}{b} \times x = \frac{a \cdot x}{b}.
\]

To prove that \( r \times x \) is irrational, we will use a proof by contradiction. Assume for the sake of contradiction that \( r \times x \) is a rational number. This means we can express it as:
\[
r \times x = \frac{c}{d}
\]
for some integers \( c \) and \( d \) (with \( d \neq 0 \)).

Rearranging gives us:
\[
a \cdot x = \frac{c}{d} \cdot b = \frac{c \cdot b}{d},
\]
which implies:
\[
x = \frac{c \cdot b}{a \cdot d}.
\]

Here, \( \frac{c \cdot b}{a \cdot d} \) is a fraction made up of integers (since \( c, b, a, d \) are all integers, and \( a \) is not zero since \( r \) is not zero). Thus, \( x \) would be expressed as a ratio of two integers, which means \( x \) is rational.

However, this contradicts our original assumption that \( x \) is irrational. Hence, our assumption that \( r \times x \) is rational must be false.

Thus, we conclude that the product of a non-zero rational number \( r \) and an irrational number \( x \) must be irrational:
\[
r \times x \text{ is irrational.}
\]

This completes the proof that a rational number multiplied by a non-zero irrational number yields an irrational product.