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My work:

  • $1.0\,000\,004 = 1+\frac{4}{10^7}=1+\frac{1}{125\cdot 10^6}$

  • $ (1.0\,000\,006)^2=(1+\frac{6}{10^7})^2=(1+\frac{3}{5\cdot 10^6})^2$

  • $ (0.9\,999\,995)^2=(\frac{9\,999\,995}{10^7})^2=(\frac{1\,999\,999}{2\cdot10^6})^2$

  • $ 0.9\,999\,998=\frac{9.999\,998}{10^7}=\frac{4\,999\,999}{5\cdot10^6}$

The two fractions become:

$A=\dfrac{1+\frac{1}{125\cdot 10^6}}{(1+\frac{3}{5\cdot 10^6})^2}$ and $B=\dfrac{(\frac{1\,999\,999}{2\cdot10^6})^2}{\frac{4\,999\,999}{5\cdot10^6}}$

At this stage, I don't know how to continue.

Any help is appreciated. Thank you.

Martin Sleziak
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Rafiq
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  • Similar question here. The answers to that question will help you solve your question. – wltrup Aug 06 '15 at 21:06
  • @ wltrup: First of all thank you. But I'm afraid i'ts not the same. – Rafiq Aug 06 '15 at 21:09
  • I didn't say it is the same question. I said it's a similar question, meaning that the methods used to solve that question are useful to solving your question. – wltrup Aug 06 '15 at 21:10

2 Answers2

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Let $a=\frac{1}{10^7}$.

Then, $$A=\frac{1+4a}{(1+6a)^2},\quad B=\frac{(1-5a)^2}{1-2a}$$ Now, $$\begin{align}A-B&=\frac{1+4a}{(1+6a)^2}-\frac{(1-5a)^2}{1-2a}\\&=\frac{(1+4a)(1-2a)-(1-5a)^2(1+6a)^2}{(1+6a)^2(1-2a)}\\&=\frac{3 a^2 (20a (1-15 a)+17)}{(1+6a)^2(1-2a)}\end{align}$$ This is positive, so $A\gt B$.

mathlove
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  • Nice and clear answer. Thank you so much. I'm not yet able to vote you. Your help is great and your idea is elegant. – Rafiq Aug 06 '15 at 21:20
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$$A=\frac {10^7\times (10^7+4)}{(10^7+6)^2}$$

$$B=\frac {(10^7-5)^2}{10^7\times (10^7-2)}$$

$$\frac AB=\frac {10^{14}(10^7+4)(10^7-2)}{(10^7+6)^2(10^7-5)^2}$$

The numerator is $10^{28}+2\cdot10^{21}-8\cdot10^{14}$

The denominator is $10^{28}+2\cdot 10^{21}-59\cdot10^{14}-60\cdot10^7+900$

It is easy to see from this which is larger.

Mark Bennet
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  • Hi, Mark Bennet. Your method is excellent. I became confident in dealing with huge numbers having exponents, and I learnt a new method of comparing two fractions, that consists of dividing the first fraction by the second. Thank you very much. – Rafiq Aug 07 '15 at 08:43