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Which is bigger: $\sqrt{1001} - \sqrt{1000}$, or $\frac{1}{10}$?

I can calculate the answer using a calculator, however I suspect to do so may be missing the point of the question.

The problem appears in a book immediately after a section called 'Rules for square roots'with $\sqrt{ab} = \sqrt{a}.\sqrt{b}$ and $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$ as the given rules.

mikoyan
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7 Answers7

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Hint: $$\sqrt{n}-\sqrt{n-1}=\dfrac{1}{\sqrt{n}+\sqrt{n-1}}$$

To prove the above,multiply the numerator and denominator of the L.H.S by the conjugate.

rah4927
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Observe the following: $$\frac{1}{10}+\sqrt{1000}=\frac{\sqrt{10}+1000}{\sqrt{1000}}>\frac{1001}{\sqrt{1000}}>\frac{1001}{\sqrt{1001}}=\sqrt{1001}$$

Spenser
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Squares of area 1000 and 1001

Is the pink border more or less than $\frac{1}{10}$cm wide?

Steve Kass
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Hint : Assume any one is bigger. write appropriate inequality sign. square. rearrange to put square root on one side. square.

evil999man
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This is not a proof but a way to "see" the answer, using just integer arithmetic. Specifically, I'll use some perfect squares near 1000: 961 and 1024 have square roots equal to 31 and 32. So an increase of 63 (from 961 to 1024) in these numbers led to an increase of only 1 in their square roots. So, if (as one might reasonably expect) square roots increase in a rather regular fashion, increasing a number near 1000 by 1 should increase its square root by roughly 1/63. Although that's only a rough estimate, it's enough to convince me that 1/10 is way too big.

Andreas Blass
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We want to find the sign of $\sqrt{1001} - \sqrt{1000} - \dfrac{1}{10}$. Multiplying by the obviously positive $\sqrt{1001} + \sqrt{1000} + \dfrac{1}{10}$ we get $(\sqrt{1001})^2 - (\sqrt{1000} + \dfrac{1}{10})^2 = 1001 - 1000 - \dfrac{2\sqrt{1000}}{10} - \dfrac{1}{100} = \dfrac{99}{100} - \dfrac{2\sqrt{1000}}{10}$

Since $\sqrt{1000} > 30$ this quantity is obviously negative. So $\dfrac{1}{10}$ is larger.

NovaDenizen
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Simply let a inequality, and see what comes up! $$\sqrt{1001} - \sqrt{1000} \gt \frac{1}{10}$$ $$(\sqrt{1001} - \sqrt{1000})^2 \gt \left(\frac{1}{10}\right)^2$$ $$1001 + 1000 - 2\sqrt{1001\cdot1000}\gt \frac{1}{100}$$ $$1001 +1000 - \frac{1}{100}\gt 2\sqrt{1001\cdot1000}$$ $$(2001 - \frac{1}{100})^2\gt (2\sqrt{1001\cdot1000})^2$$ $$2001^2 + \frac{1}{100^2}-2\frac{2001}{100}\gt 4\cdot1001\cdot1000$$ $$(1000+1001)^2 + \frac{1}{100^2}-2\frac{2001}{100}\gt 4\cdot1001\cdot1000$$ $$1000^2+1001^2+2\cdot10001\cdot1001 + \frac{1}{100^2}-2\frac{2001}{100}\gt 4\cdot1001\cdot1000$$ $$1000\cdot1000+1001^2+\frac{1}{100^2}-2\frac{2001}{100}\gt 2\cdot1001\cdot1000$$ $$1001^2+\frac{1}{100^2}-2\frac{2001}{100}\gt 1000(2002-1000)$$ $$1001^2+\frac{1}{100^2}-2\frac{2001}{100}\gt 1000\cdot1002$$ $$(1000+1)^2+\frac{1}{100^2}-2\frac{2001}{100}\gt 1000\cdot1002$$ $$1000^2+1+2000+\frac{1}{100^2}-2\frac{2001}{100}\gt 1000\cdot1002$$ $$1+2000+\frac{1}{100^2}-2\frac{2001}{100}\gt 1000\cdot(1002-1000)$$ $$2001+\frac{1}{100^2}-2\frac{2001}{100}\gt 2000$$ $$1+\frac{1}{100^2}\gt 2\frac{2001}{100}$$ $$100^2+1\gt 2\cdot 2001\cdot100$$ $$1\gt 100(4002-100)$$ $$1\gt 100\cdot3902$$ which is evidently wrong. It couldn't be neither $$\sqrt{1001} - \sqrt{1000} = \frac{1}{10}$$ as you would come up with $$1= 100\cdot3902$$ which is false too... so $\sqrt{1001} - \sqrt{1000} \gt \frac{1}{10}$ is false, then $$\sqrt{1001} - \sqrt{1000} \lt \frac{1}{10}$$

sirfoga
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  • Or possibly $\sqrt{1001}-\sqrt{1000}=1/10$, but that is pretty easy to dismiss by a rational/irrational argument. – Gamma Function Apr 21 '14 at 17:23
  • yes, sorry, I took it for granted... – sirfoga Apr 21 '14 at 17:24
  • Yuck! Compare this with rah4927's answer. – TonyK Apr 21 '14 at 19:16
  • Disagree, this is mostly fine, "square both positive sides until there is no longer a radical" is probably the best way to solve this. It's a very general approach that doesn't require an epiphany. He just takes too long to get to the point after the radicals are removed, could just multiply both sides by $100^2$ and be done with it. – DanielV Apr 21 '14 at 19:31
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    @DanielV: "Probably the best way to solve this" is just to note that $(\sqrt{1001}-\sqrt{1000})(\sqrt{1001}+\sqrt{1000}) = 1$. That's all you need, and it generalises to many similar problems. – TonyK Apr 21 '14 at 19:54
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    I thought it was rather simple and school-like answer... – sirfoga Apr 22 '14 at 15:40