2
$\begingroup$

I wanted to use this fact in a proof, it seems obvious, but I should probably prove this as well. Any proof will do. Also for the future is there a way to search for these proofs google does not work very well.

EDIT: Ah, it may also be good to show that the n-th root of any real number a s.t. $0\lt a\lt1$ is also $\lt1$

Thanks.

  • 7
    Can you prove that if $0 < b < 1$ then $b^2 < b$?2010-11-11
  • 2
    Regarding your edit: Yes, it wouldn't hurt to show that. What have you tried?2010-11-11

5 Answers 5

6

If you know that whenever $a\lt b$ and $c\gt 0$ then $ac\lt bc$, then for any $x$ with $0\lt x\lt 1$, multiplying through by $x$ gives \begin{equation*} 0 \lt x^2 \lt x \lt 1. \end{equation*} (The last inequality is "inherited" from the fact that you already know that $x\lt 1$). Repeat to get \begin{equation*} 0\lt x^3 \lt x^2 \lt x \lt 1. \end{equation*} And so on; you get $0\lt x^n \lt x^{n-1}\lt\cdots \lt x \lt 1$.

Working the other way, if $1\lt y$, then multiplying through by $y$ you get \begin{equation*} 1\lt y\lt y^2 \end{equation*} (with the first inequality "inherited from your original assumption) and repeating this process leads to \begin{equation*} 1\lt y\lt y^2 \lt\cdots \lt y^n \end{equation*} So: if $0\lt z\lt 1$, then $0\lt z^n\lt 1$. If $1\lt z$, then $1\lt z^n$. And of course, if $z=1$ then $z^n = 1$.

So, start with $0\lt a\lt 1$. Then $\sqrt[n]{a}$ cannot be greater than $1$ (then $a = \left(\sqrt[n]{a}\right)^n \gt 1$ by the above), so $\sqrt[n]{a}\lt 1$.

Now replace $x$ with $\sqrt[n]{a}$ to get the desired inequality.

  • 1
    Don't you first need to show $\sqrt[n]{a} \le 1$?2010-11-11
  • 1
    @Moron: Well, okay; you need to know that if $x\lt 1$ then $x^n\lt 1$, and if $x\gt 1$ then $x^n \gt 1$. You've essentially already shown it, but I'll add it.2010-11-11
  • 1
    Yeah, just nitpicking :-)2010-11-11
6

HINT $\rm \ \sqrt a - a \ =\ \sqrt a\ \:(1 - \sqrt a)\ > 0\ $ since both factors are $> 0$ on the interval $(0,1)$.

3

We argue by contradiction. Suppose $a \in (0,1)$ and $\sqrt{a} \leq a$. Then \begin{eqnarray} \frac{1}{\sqrt{a}} = \frac{\sqrt{a}}{a} \leq 1, \end{eqnarray} which implies that $\sqrt{a} \geq 1$ and therefore $a \geq 1$, which is absurd by construction.

2

Depending on the level of the audience, it might be seen as obvious. I would just argue that $\sqrt{x}$ is monotonic and $\gt 0$ on $(0,1)$, so $\sqrt{a}\lt1$ then multiply both sides by $\sqrt{a}$

0

In a simply way we have $a<1$.

Multiply both side by $a$ then we get $a^2

Taking the square root for both sides we get $a < \sqrt{a}$.