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Show that $p^2=2$ is not satisfied by a rational number
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We now show that the equation
$p^{2}=2 \dots (1)$
is not satisfied by any rational $p$. If there were such a $p$, we could write $p=m / n$ where $m$ and $n$ are integers that are not both even. Let us assume this is done. Then (1) implies
$m^{2}=2 n^{2} \ldots (2)$
This shows that $m^{2}$ is even. Hence $m$ is even (if $m$ were odd, $m^{2}$ would be odd), and so $m^{2}$ is divisible by 4 . It follows that the right side of (2) is divisible by 4 , so that $n^{2}$ is even, which implies that $n$ is even.

The assumption that (1) holds thus leads to the conclusion that both $m$ and $n$ are even, contrary to our choice of $m$ and $n$. Hence (1) is impossible for rational $p$.

We now examine this situation a little more closely. Let $A$ be the set of all positive rationals $p$ such that $p^{2}<2$ and let $B$ consist of all positive rationals $p$ such that $p^{2}>2$. We shall show that $A$ contains no largest number and $B$ contains no smallest.

More explicitly, for every $p$ in $A$ we can find a rational $q$ in $A$ such that $p<q$, and for every $p$ in $B$ we can find a rational $q$ in $B$ such that $q<p$.
To do this, we associate with each rational $p>0$ the number

$q=p-\frac{p^{2}-2}{p+2}=\frac{2 p+2}{p+2} \ldots (3)$
Then
$q^{2}-2=\frac{2\left(p^{2}-2\right)}{(p+2)^{2}} \ldots (4)$
If $p$ is in $A$ then $p^{2}-2<0,(3)$ shows that $q>p$, and (4) shows that $q^{2}<2$. Thus $q$ is in $A$.

If $p$ is in $B$ then $p^{2}-2>0,(3)$ shows that $0<q<p$, and (4) shows that $q^{2}>2$. Thus $q$ is in $B$.
$1.2$ Remark The purpose of the above discussion has been to show that the rational number system has certain gaps, in spite of the fact that between any two rationals there is another: If $r<s$ then $r<(r+s) / 2<s$. The real number system fills these gaps. This is the principal reason for the fundamental role which it plays in analysis.
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