MATH.21:5 - Archetypal Grounding

MATH.21:5.1 - Obtain a real object from rational intervals

The task is to construct a positive number r with r²=2 and obtain rational approximations with a chosen absolute error.

Start with l_0=1 and u_0=2. At each step take the rational midpoint m. If m²≤2, replace the lower endpoint by m; otherwise replace the upper endpoint. Squaring is increasing on the positive interval, so each step retains l_n²≤2≤u_n². The intervals are nested and their widths are 2⁻ⁿ.

The lower endpoints form a bounded increasing sequence. In the real numbers, their supremum r exists. Each l_n≤r≤u_n: every later lower endpoint is at most u_n, and earlier ones are no larger than l_n. The shrinking width makes this r the only common point.

Both r² and 2 lie between l_n² and u_n². Since the endpoints stay between 1 and 2, the width of this squared interval is (u_n-l_n)(u_n+l_n)≤4·2⁻ⁿ. It tends to zero, so r²=2. This obtains the desired object without presupposing a square-root value to drive the construction.

After four bisections the interval is [22/16,23/16]. Its midpoint 45/32 differs from r by at most 1/32. For a smaller tolerance epsilon, choose a stage with 2⁻ⁿ⁻¹≤epsilon, or refine until the interval width is at most twice epsilon.

Changed space. If the result must remain rational, existence fails. In a fraction p/q in lowest terms, p²=2q² would make p even, and then q even, contradicting lowest terms. The rational approximations and their widths remain available, but they construct a real number rather than a missing rational solution. The next decision concerns admitting that extension or retaining a finite rational answer.

MATH.21:5.2 - Construct an infinite word and return a finite observation

Let p_n be a binary word of length n, with p_n a prefix of p_(n+1). Positions start at zero. Define b(k) to be entry k of p_(k+1). Compatibility makes that entry agree with every later prefix, so b is an infinite binary word whose first n entries are p_n.

For a common space of approximations, pad each p_n with zeros to obtain an infinite word b_n. Use convergence by eventual agreement on each finite prefix. For a request about the first m entries, every b_n with n≥m agrees there with b. This proves convergence and gives the return stage: p_m supplies those entries. Any calculation depending only on them, such as their count of ones, can use that finite result.

The global property “has only finitely many ones” does not pass to this limit. Take p_n to consist of n ones. Each zero-padded b_n has finitely many ones, but b has a one at every position. A question about a fixed finite prefix is settled; a question about the entire tail needs another argument.

The construction uses a consistency relation between finite stages and a rule for obtaining every entry. Its computational realization additionally needs a way to obtain the required p_m. Merely knowing that such prefixes exist does not supply their generator.

MATH.21:5.3 - Preserve values and integrals, then inspect differentiation

For n≥1, let f_n(x)=sin(nx)/n on [-1,1]. The inequality |f_n(x)|≤1/n holds for every input in the interval. Thus the functions converge uniformly to f(x)=0.

For a requested value error epsilon, n≥1/epsilon is sufficient. Integration over the interval is also controlled: the absolute integral of f_n-f is at most 2/n, by the interval length and the uniform bound. This estimate supplies an integral-error result without depending on cancellation of positive and negative values.

Now ask for the derivative at zero. Each f_n has derivative cos(nx), so f_n’(0)=1 for every n. The limit function has f’(0)=0. The value convergence therefore fails to justify passing this operation through the limit, even at that one point.

One possible sufficient repair, for differentiable functions on a common open interval, is pointwise convergence of the functions together with locally uniform convergence of their derivatives. A suitable limit theorem then identifies the limiting derivative. The present f_n fail that condition. If the construction can change, a family with controlled derivatives can support the stronger use; if it must stay fixed, obtain the derivative of its limit by another argument.

The result retains uniform convergence and its finite value and integral bounds. Only the derivative interchange has been refused. This is why the requested next operation belongs in the initial choice of approximation.