# Why the square root of any decimal number between 0 and 1 always come out to be greater than the number itself?

Why the square root of any decimal number between 0 and 1 always come out to be greater than the number itself? Whereas if we take the square root of say 25 we are left with 5, which is less than the number 25.

• Deleting my comment, because user296602 describes the same idea in their answer, posted at more or less the same time :-) (+1). – Jyrki Lahtonen Jan 23 '18 at 20:07

## 8 Answers

It's because if $t \in (0, 1)$ then $t^2 < t$. To see this, just remember that

$$t^2 = t \times t < 1 \times t = t$$

for these values. Now if we think of $t = \sqrt{x}$, then we have the inequality $x < \sqrt{x}$.

This same argument works, of course, for $t > 1$ with the opposite inequality. So it explains what you've noticed for $x > 1$ too.

• That is nice and simple. – Mark Bennet Jan 23 '18 at 20:00
• This is good... – Randall Jan 23 '18 at 20:05
• I think you should clarify your notation "$(0,1)$". It could be misleading in that it could be a coordinate on a Cartesian plane or the greatest common divisor etc. What is your set here (I am missing the "where")? I am thinking it should be $∀t \in ℕ | 0 ≤ t ≤ 1$ – Jonathan Komar Oct 29 '18 at 6:37
• @JonathanKomar (0, 1) is standard notation for the open unit interval, and the context implies that it's not something in the plane or a gcd. It's also not the set $\{t \in \mathbb{N} : 0 \le t \le 1\}$, because that set has at most two elements (neither of which satisfy $t^2 < t$.). – user296602 Oct 29 '18 at 21:07
• Simple mistake. $\{∀t \in ℝ | 0<t<1\}$. You're right about it being standard, but it overlaps with quite a few other standards, that's the thing. – Jonathan Komar Oct 29 '18 at 23:55

I'll try. Here are graphs of $y=x^2$ (red) and $y=x$ (blue).

You can see that when $x$ is properly between $0$ and $1$ that we have $x^2 < x$. Here, $x$ is your square root of $x^2$, and sure enough, it is larger.

To rephrase my ending, when you square "large" numbers they get larger. When you square "small" numbers they get smaller. If you turn this around to be about square roots instead, you get your question.

• A picture is worth a thousand words :) – Yuriy S Jan 23 '18 at 19:49

I know that it seems counterintuitive. The product of two numbers, each bigger than one, is bigger than either of them. However, the reciprocal of a number bigger than one is between zero and one and vice versa. But the reciprocal operation reverses the order relations. For example, two is less than three, but one half is bigger than one third. Thus, the square of a number bigger than one is bigger than the original number, and therefore, the square root is less than the original number. Taking reciprocals, the order relation is now reversed.

A similar situation is the case that the sum of two positive numbers is bigger than either of them. However, taking the negative of a number reverses the order relations. For example, two is less then three, but negative two is bigger than negative three. And so on. This may be helpful for you to think about.

Think about a decimal number between 0 and 1 as a fraction with its numerator GREATER than its denominator. Say you are taking the square root of the number $1/25$. So, you acquire $\sqrt{1/25}$ as the expression which you have to evaluate. This becomes $\sqrt{1}/\sqrt{25}$, or $1/5$. $1/5 > 1/25$.

If you still do not understand, just take a random decimal, say $x$ with possible values $0<x<1$. Turn your decimal into a fraction. Now, multiply the numerator $n$ by its multiplicative inverse to turn the numerator into 1. The denominator $d$ becomes a certain number, so the expression will look like $1/d$. You are actually making the denominator smaller by square rooting it, but as the denominator gets smaller, the final value of your expression gets larger.

Edit: I know this is 6 hours late, but a student in elementary school would understand this. Take your time to grasp the reasoning.

If you multiply any positive number by a positive number less than $1$, you make it smaller.

So since $x<1$, you get $x^2<x$

$$\left(\sqrt{x}-x\right)'=\frac{1}{2\sqrt{x}}-1>0 \Leftrightarrow \sqrt{x}<\frac{1}{2} \Leftrightarrow x<\frac{1}{4}$$

Hence the function is increasing on $\displaystyle \left[0,\frac{1}{4}\right]$ and decreasing on $\displaystyle \left[\frac{1}{4}, +\infty\right[$. It is null where $x=0$ and $x=1$ hence it is positive on $\left[0,1\right]$ then negative en $\left[0,+\infty\right[$. Here's why.

One way to see this is to consider $f(x)=x^2-x=(x-\frac 12)^2-\frac 14$ (completing the square)

$f(x)=0$ for $x=0, x=1$ (and as a quadratic, these are the only zeros) and has a minimum value at $x=\frac 12$ with $f(\frac 12)=-\frac 14\lt 0$ so $f(x)\lt 0$ for $x\in (0,1)$ ie $x^2\lt x$

For $x\gt 1$ or $x\lt 0$ we find that $f(x)\gt 0$ whence $x^2\gt x$

Because when you multiply 2 numbers between 0, and 1. The answer is smaller, that is why when u take square root of a number between 0, and 1, The answer is bigger.

• Please take the time to review this problem to fix grammatical errors. – Yash Jain Jan 24 '18 at 13:07