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Explain why $9<\sqrt{89}<10$. How do you explain this? I'm doing revision and we haven't been taught it yet but it will be on the test.

$\sqrt{389}$ is also between two consecutive whole numbers. What are the two numbers?

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The square root is an increasing function so $x<y\Rightarrow \sqrt{x}\lt \sqrt{y}$. $81\lt 89\lt 100$ so $\sqrt{81}\lt \sqrt{89}\lt \sqrt{100}\Rightarrow 9\lt \sqrt{89}\lt 10$. –  user69810 May 8 '13 at 19:25
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For the second part, did you mean "between two consecutive square numbers"? –  Jerry May 8 '13 at 19:28
    
The second part is just "Counting". I presume you're intelligent enough to perceive what comes before 389 and what follows it. –  Maazul May 8 '13 at 19:29
    
Yeah it's 19 and 20. –  user61406 May 8 '13 at 19:29
    
Right. $19^2$ gives $361$ and $20^2$ gives $400$. –  Jerry May 8 '13 at 19:30

9 Answers 9

up vote 13 down vote accepted

Hint:

You want to demonstrate -

$9<\sqrt{89}<10$

What happens when you square across these terms?

This can be done because, based on the suggested edit by Cameron Buie, each term can be squared for an equivalent chain of inequalities since $x↦x^2$ is a strictly increasing function on the positive reals.

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81 < 89 < 100, yeah I got it 89 is in between 81 and 100 of course –  user61406 May 8 '13 at 19:27
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-1...therefore we have $1<-2$ because we can square them and get that $1<4$... –  Jp McCarthy May 9 '13 at 17:41
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@JpMcCarthy, First your downvote seems to be too rough, even in the answerer made a mistake, which is not sure. Second, it could be argued that the answer's argument is fine here since everything's positive and even bigger than 1... –  DonAntonio May 9 '13 at 19:33
    
@JpMcCarthy You are correct - I should had been more careful in being sure to specify the domain; though I do agree with DonAntonio that a (-1) is a bit excessive and that just a comment with counter-example sufficient. –  Jordan Mahar May 9 '13 at 20:04
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Apologies. I will remove the downvote when I can. My frustration was more with the abundance of upvotes rather than the answer itself. I think Sami's answer is the best and will upvote that rather than be negative about another answer. I suppose seeing so many students thinking that $A\Rightarrow B$ means that $B\Rightarrow A$ also has made me battle-weary! –  Jp McCarthy May 9 '13 at 22:30

Recall that the square root is monotonically increasing function so $$81<89<100\Rightarrow\sqrt{81}=9< \sqrt{89}<10=\sqrt{100}$$

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In addition to the excellent answers above, I'd like to add this picture. (My tikz-fu needs all the practice I can give it).

enter image description here

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i trust in wolfram too (+1) –  Mula Ko Saag May 13 '13 at 18:07

Hint: what's $9^2$, and what's $10^2$?

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I got it, it's $19^2=361$ and $20 =400$ for the two consecutive numbers on both sides of $\sqrt{389}$.

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Imagine a triangle (with one corner90) (with one corner( not 90 degree) in (0,0)point ) and one edge=$\sqrt8$ then other edge=9(on HORIZONTAL axis ) then if you imagine a CIRCLE with center (0,0)and $r=\sqrt{89}$ you will see that length of SINEW of triangle ($\sqrt{89}$)will be larger than length of edge=9.

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Hints:

$$9^2=81<89<100=10^2\ldots$$

Of course, you may also want to show the square root function $\,f(x):=\sqrt x\,$ is monotone ascending, for example by means of its first derivative....

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$f(x)=x^2$ is a strictly increasing function on the positive real numbers, meaning that $$a<b\Leftrightarrow f(a)<f(b) \hspace{30pt} \forall \hspace{5pt} 0\leq a,b\in\mathbb{R}.$$ That means that it suffices to notice that $$\color{darkblue}{9^2=81}<\color{darkred}{89}<\color{purple}{100=10^2}$$ which therefore implies that $$\color{darkblue}{9}<\color{darkred}{\sqrt{89}}<\color{purple}{10}.$$ Here is some visual intuition on why the algebra works. The thick blue represents the graph of the function $f(x)=x^2$.

enter image description here

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$$\begin{align*} 9<\sqrt{89}<10&\Rightarrow \sqrt{9^2}<\sqrt{89}<\sqrt{10^2}\\\,\\&\Rightarrow\sqrt{81}<\sqrt{89}<\sqrt{100}\,\,\,\checkmark \end{align*}$$

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