Question

7. Prove that for any positive real number α there exists a sequence of real numbers converging to √

α.​

Answer 1

Answer:

Proof given below:

Step-by-step explanation:

To prove:

for every real number a there exists a sequence $r_n$ of rational numbers such that $r_n$→$a$.

We can suppose that the real number an is greater than zero without losing generality. (If a0, we can use the following argument on |a| and then switch signs.) Based on the fact that the reals constitute a complete ordered field, we sketch a rather rigorous proof. We provide an easy informal but incomplete "evidence" in one of the closing remarks.

Let $n$ be a natural number. Let $m=m(n)$ be the largest positive integer such that $\\{ {m} \atop {n}$$\geq a$

Let  $r_n$$=m/n$. The sequence ($r_n$) has limit $a$, as can be seen from the definition of limit.

1. It is necessary to prove that there is a positive integer m such that

$\\{ {m} \atop {n}$$\geq a$. It is sufficient to demonstrate that there is a positive integer k such that ka, and then we can use m=kn. The Archimedean property of the reals refers to the notion that there is always an integer >a. We next proceed to demonstrate that the reals own this property.

Assume that all positive integers are 0 and that there is an integer k such that 0b, which contradicts the premise that b represents an upper bound for N.

2. One may also provide a rapid, but insufficiently convincing "proof" of the approximation result. Assume that a>0 as before. The numbers formed by truncating a's decimal expansion at the n-th place are rational and have limit a. The issue is because we're presuming that every real number has a decimal extension at this point.