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DeanR

Great question.  Let's let r be a rational number and s be irrational.  Note r has to be nonzero for this to work.   In other words, it's not true that when we multiply zero, a rational number, by an irrational number like π we get an irrational number.  We of course get zero.

The question is: why is the product

[tex]p = rs[/tex]

irrational?

In math "why" questions are usually answered with an illuminating proof.  Here the indirect proof is enlightening.

Suppose p was rational.   Then

[tex]s = \dfrac p r[/tex]

would be rational as well, being the ratio of two rational numbers, so ultimately the ratio of two integers.  

But we're given that s is irrational so we have our contradiction and must conclude our assumption that p is rational is false, that is, we conclude p is irrational.


konrad509

A proof by contradiction.

Let assume that the product of a rational number and an irrational number is rational.

Let [tex]\dfrac{a}{b}[/tex] and [tex]\dfrac{c}{d}[/tex] be rational numbers, where [tex]a,b,c,d\in \mathbb{Z} \wedge b,d\not=0[/tex] and [tex]x[/tex] an irrational number.

Then

[tex]\dfrac{a}{b}\cdot x=\dfrac{c}{d}\\ x=\dfrac{bc}{ad}[/tex]

Integers are closed under multiplication, therefore [tex]bc[/tex] and [tex]ad[/tex] are integers, making the number [tex]x=\dfrac{bc}{ad}[/tex] rational, which is contradictory with the earlier statement that [tex]x[/tex] is an irrational number.

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