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Thread: infimum and supremums

  1. #1
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    infimum and supremums

    We showed in class that sqr root 2 is not rational, i.e., that there

    is no rational number y such that y^2= 2. The aim of this

    problem is to show that there exists a positive real number

    y such that y^2= 2.

    Let A = {x is in R | x^2< 2}.

    (a) Show that A is subset of [−2, 2].

    (b) Suppose that y is the least upper bound of A. Show

    that

    i) y >= 1;

    ii)y^2 >= 2;

    iii)y^2 <= 2;

    iv)y^2 = 2.

    [Hint: Use proofs by contradiction for parts ii)–iii).]
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by alexmin View Post
    We showed in class that sqr root 2 is not rational, i.e., that there

    is no rational number y such that y^2= 2. The aim of this

    problem is to show that there exists a positive real number

    y such that y^2= 2.

    Let A = {x is in R | x^2< 2}.

    (a) Show that A is subset of [−2, 2].
    we denote the solution to $\displaystyle a^2 = 2$ to be $\displaystyle a = \pm \sqrt{2}$. then clearly, for all $\displaystyle x \in A$, $\displaystyle x \in (-\sqrt {2}, \sqrt {2})$

    it can be shown (by examples if necessary) that $\displaystyle |\sqrt{2}| < 2$. thus $\displaystyle A = (-\sqrt{2}, \sqrt{2}) \subset [-2,2]$


    (am i begging the question here? i hate doing proofs for things that seem "obvious")


    (b) Suppose that y is the least upper bound of A. Show

    that

    i) y >= 1;
    well, if you accept what is in (a), this is immediate. i guess you can show it by example, or come up with some approximation to $\displaystyle \sqrt{2}$ or just some number greater than 1 but less than $\displaystyle \sqrt{2}$, and then it would be in A.

    ii)y^2 >= 2;
    assume to the contrary that y^2 < 2. then by the denseness of Q, we can find some rational x in A such that y^2 < x^2 < 2. a contradiction

    iii)y^2 <= 2;
    assume to the contrary that y^2 > 2. clearly we have that y^2 > 2 > x^2 for all x in A. but that means that there is some upper bound of the set A that is less than y, therefore, y cannot be the supremum as stated. there's our contradiction

    iv)y^2 = 2.
    i'm not in the mood for any fancy arguments at this point, i never liked this question to begin with. so i'd just slap on the abstract property of real numbers that says: if a and b are real numbers, with a <= b and b <= a then a = b
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  3. #3
    Senior Member JaneBennet's Avatar
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    (b)(i) is straightforward: use the definition of upper bound. $\displaystyle y$ is an upper bound of $\displaystyle A\ \Rightarrow\ \forall\,x\in A, x\leq y$. In particular, $\displaystyle 1\leq y$ since $\displaystyle 1\in A$.

    b(ii)
    If $\displaystyle y^2<2$, we can find a real number $\displaystyle x$ such that $\displaystyle y^2<x<2$. Since $\displaystyle x$ is positive, we can write $\displaystyle x=a^2$ for some positive real number $\displaystyle a$. Thus $\displaystyle y^2<a^2<2$. But $\displaystyle a^2<2\ \Rightarrow\ a\in A\ \Rightarrow$ (given that $\displaystyle y$ is an upper bound of $\displaystyle A$) $\displaystyle a\leq y\ \Rightarrow\ a^2\leq y^2$. That’s our contradiction.

    b(iii)
    If $\displaystyle y^2>2$, take a real number $\displaystyle x$ such that $\displaystyle 2<x<y^2$. As before we can write $\displaystyle x=a^2\ (a>0)$, so $\displaystyle 2<a^2<y^2$. $\displaystyle a^2<y^2\ \Rightarrow\ a<y\ \Rightarrow\ a$ is not an upper bound of $\displaystyle A$ since $\displaystyle y$ is the least upper bound $\displaystyle \Rightarrow\ \exists\,b\in A$ such that $\displaystyle a<b$. But $\displaystyle b\in A\ \Rightarrow\ b^2<2$; then $\displaystyle a<b\ \Rightarrow\ a^2<b^2<2$. Contradiction again.

    Note that b(ii) is true for all upper bounds of $\displaystyle A$, not just the least upper bound. b(iii) only holds for the least upper bound.
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