Notes on the family S2

MATH

  1. Proposition: $S2$ is consistent.

    Proof

    Suppose $[r,s]$ and $[u,v]$ are intervals in $S2$. We must prove $r\leq v$ and $u\leq s$. Suppose $r>v$. The we have $r^{2}>v^{2}\geq 2$, so $r^{2}>2$, contradicting $r\leq 2$. We can show in a similar fashion that $u>s$ is impossible.

  2. Proposition: $S2$ is a fine family.

    Proof

    The interval $I_{0}=[1,2]$ lies in $S2$ since MATH. We construct other intervals $I_{k},\ k=1,2,...,$ in the following way. Suppose $I_{k}$ has been constructed (e.g. $I_{0}$). Let MATH, and let MATHmidpoint of $I_{k}$. If $m^{2}\geq 2$ then $I_{k+1}=[r_{k},m]$; if $m^{2}<2$, the $I_{k+1}=[m,s_{k}]$. It is each to check that $I_{k+1}$, defined in this way, belongs to $S2$, and that MATH. Given $\epsilon >0$, we can choose $k$ so that $2^{k}>1/\epsilon $; for such a $k$, MATH. (For example, if $\epsilon =1/1000$, then MATH.) Thus, $S2$ contains arbitrarily fine intervals.

  3. Proposition: If $x$ belongs to every interval of $S2$, then $x^{2}=2$.

    Proof

    We will show that MATH for every $\epsilon >0$. Let $[r,s]$ be any interval in $S2$. Since $x$ belongs to $[r,s]$, MATH; since $[r,s]$ belongs to $S2$, MATH. Thus, both $x^{2}$ and $2$ belong to $[r^{2},s^{2}]$. Now MATH. Since $x$ belongs to every interval in $S2$, we can assume that $[r,s]$ is one of the intervals constructed by bisection in the previous Proposition. If this is the case, $1\leq r\leq 2$ and $1\leq s\leq 2$, so $(s+r)\leq 4$, and we have MATH. Since the intervals constructed by bisection are fine, choose $[r,s]$ so that $s-r<\epsilon /4$; then MATH. Thus, $x^{2}$ and $2$ lie in an interval of length $<\epsilon $, so MATH for any $\epsilon >0$. By the "wiggle" lemma, MATH, so MATH, and so $x^{2}=2$.

  4. Proposition: No rational number can have $2$ as its square.

    Proof (Adapted from Euclid)

    Suppose $r=\dfrac{A}{B}$ has $2$ as its square. By repeated cancellation of $2$s from numerator and denominator, we can replace $A$ and $B$ by whole numbers $M$ and $N$ which are not both even; Thus, we have MATH. Therefore $M^{2}=2N^{2}$ and so $M^{2}$ is even. Now $M$ can't be odd or its square would be odd (odd times odd = odd), so $M$ must be even; $M=2K$. Substituting this into the previous equation gives: MATH, so $2K^{2}=N^{2}$. Therefore $N^{2}$ is even, so $N$ must be even also. Thus, contrary to assumption, $M$ and $N$ are both even. This contradiction shows that the assumption that a rational number has square $2$ is untenable.

  5. Proposition: As a real number, MATH.

    Proof

    We must show that, as families, MATH. This amounts to showing that every interval $[r,s]$ in $(S2)^{2}$ meets $[2,2]$; i.e. contains $2$. A typical interval in $(S2)^{2}$ looks like MATH where $[r,s]$ and MATH belong to $S2$. We haveMATHThus, MATH and we are done: $(S2)^{2}=2$.

  6. From the above it follows that $S2$ is a real number whose square is $2 $, so we are justified in actually calling it $\sqrt{2}$. Furthermore, $S2$ is a consistent family of rational intervals, but there is no single rational number which lies in all of its intervals. (If there were, it would be a rational with $2$ as its square, by part 3 above. But this is impossible, by part 4.)