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.
2026-02-22 19:53:44.1771790024
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Why the square root of any decimal number between 0 and 1 always come out to be greater than the number itself?
12.4k Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
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There are 10 best solutions below
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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.
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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.
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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 $x^2\gt x$
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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.
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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.
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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.
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If you multiply any positive number by a positive number less than $1$, you make it smaller.
So since $x<1$, you get $x^2<x$
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it’s just the decimal numbers below $1$ whose value gets decreased because you are actually multiplying them by a number less than one thus decreasing it’s value. for eg . $0.5 \cdot 0.5$ , here you are multiplying $0.5$ by $0.5$ , i.e. you are halving the number again.
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I could see most of the above answers give a clear idea that how $\sqrt x > x \text{ if } x \in (0, 1)$ but I would like to answer what I had explained to my $9$yrs old sister.
I asked her what did you mean by a number less than $1$?
Ans: Something in decimal like $0.5$
So, We can write it as $ \frac {5}{10} = \frac 1{2}$ or in a form of $\frac {\text{smaller}}{bigger}$ and when we square this number $(\frac 1{5})^2 = \frac {1^2}{5^2} = \frac {1}{25}$
This means first we were dividing our number $1$ by $5$ but now after squaring our number $1$ has been divided by the square of $5^2 = 25$ Thus, our number $x$ has become smaller than it was before we squared.
$x^2 < x \text{, }\forall x\in (0, 1)$
And which we later stated as: The square of X becomes less than X if X $\in(0, 1)$ e.i Square-root of X would be greater than X if $X \in (0, 1)$
That is how I could try to persuade a child!
- How we can approach $$\text{smaller}<\text{bigger} \implies \frac {smaller}{bigger} < 1 \implies \frac {smaller\times bigger}{bigger\times bigger} > \frac {smaller\times \color{red}{smaller}}{bigger \times bigger} \implies x>x^2$$
In the end, a would use derivatives.
$$ \left(\sqrt{x}-x\right)'=\frac{1}{2\sqrt{x}}-1>0 \Leftrightarrow \sqrt{x}<\frac{1}{2} \Leftrightarrow x<\frac{1}{4} $$
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.