I was recently working on a question essentially worded in the following way:
Where does a proof of $\sqrt{4}$ being irrational fall apart when we try to apply the same principles used for proving that $\sqrt{2}$ is irrational.
I attempted by making the same (in this case, intuitively correct) assumptions that led to a contradiction in the case of $\sqrt{2}$:
$\sqrt{4}$ is a rational number and can be written as $\dfrac{m}{n}$ where $n\neq0$
$\dfrac{m}{n}$ is in lowest reduced terms; i.e. $m$ and $n$ are co-prime due to definition of rational numbers
Then I took the following steps:
$$m^2 = 4n^2$$
$$m^2 = 2(2n^2)$$
Thus, $m^2$ is even $\implies$ $m$ is even and can be written as $2k$.
$$m^2 = 4k^2 = 4n^2$$
$$k = n$$
Thus, $k$ is a factor of both $m$ and $n$, contradicting the second assumption that I made ($m$ and $n$ are co-prime).
Although I understand intuitively that this is not the case, doesn't this show that $\sqrt{4}$ is irrational?
Answer
You have proven that $n = k$ and $m = 2k$. In the case that $m$ and $n$ are coprime, set $k = 1$.
No comments:
Post a Comment