Proof of the existence of the square root of positive numbers

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Well, suppose we would have $a < x$. Then there would be some number $\delta > 0$ such that $x - a = \delta$, or $x = a + \delta$. Plugging this into your $a + e(1 + 2n) > x$ yields $$ x= a + \delta < a + e(1 + 2n) \\ \delta < e(1 + 2n) $$ But remember that we can make $e$ arbitrarily small, whereby $\delta$ would be a positive number, smaller than every positive number. Such a number cannot exist.

Hence $a < x$ is impossible, so we must have $a \geq x$.

Another way to look at it: the set of all numbers of the form $a + e(1 + 2n)$ is bounded from below by $x$, so it must have a highest lower bound $l$. Then $x \leq l \leq a$ (any higher number than $a$ cannot be a lower bound).

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Updated on January 15, 2020

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  • ObnoxiousFrog
    ObnoxiousFrog almost 4 years

    For any $x \ge 0$ there exists $y \ge 0$ and $y^2=(−y)^2=x$. So for any $y \in \Bbb R$ with $x \ge 0$ we have $\sqrt y \in \Bbb R$.

    Partial Solution: Let $A = \{z \in \Bbb R : z^2 < x\}$. It is a bounded set. By the Archimedean property there exists an $n \in \Bbb N$ such that $n > x$. So by the completeness axiom, there exists a least upper bound to the set, call this $y$. Let $y^2=a$. Our aim is to show $a = x$.

    Let $1 > \varepsilon > 0$ and choose $n \in \Bbb N$ with $n \ge y$ and note that

    $$(y + \varepsilon)^2 = y^2 + 2 y \varepsilon + \varepsilon^2 \le a + \varepsilon (1+2y) \le a + \varepsilon (1+2n) .$$

    Since $(y + \varepsilon)^2 > x$ implies $a + \varepsilon (1+2n) > x$, if we take $\varepsilon$ to be arbitrarily small this means $a \ge x$.

    My question:

    Why does $a + \varepsilon (1+2n) > x$ imply $a \ge x$? Is this similar to saying $2 \ge 1$?

    • vrugtehagel
      vrugtehagel almost 8 years
      I suppose you mean $\mathbb{R}$ - try writing \mathbb{R}.
    • Alex M.
      Alex M. almost 8 years
      The sentence "So for any $y \in \Bbb R$ with $x \ge 0$ we have $\sqrt y \in \Bbb R$." makes absolutely no sense. You must have miscopied (and misunderstood) it.
    • ObnoxiousFrog
      ObnoxiousFrog almost 8 years
      Might be a typo from my lecture notes but its word for word :/
    • Bib-lost
      Bib-lost almost 8 years
      I think Alex M. is suggesting it should be $y \geq 0$ instead of $x$.