Albert1
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compare :
$7$, and $\sqrt 2+\sqrt 5 + \sqrt {11}$
which one is bigger?
$7$, and $\sqrt 2+\sqrt 5 + \sqrt {11}$
which one is bigger?
The discussion revolves around comparing the numerical value of $7$ with the expression $\sqrt{2} + \sqrt{5} + \sqrt{11}$. Participants are exploring methods to determine which is larger without the use of a calculator.
The discussion does not reach a consensus on which value is larger, and participants have not settled on a definitive method or solution.
Participants have not fully detailed their proposed methods or the assumptions underlying their approaches, leaving some steps and reasoning unresolved.
now ,try to prove it,without using calculatorgreg1313 said:Approximating the roots to 3 significant digits, we have$$\sqrt2+\sqrt5+\sqrt{11}\approx1.41+2.23+3.31=6.95,6.95+3\cdot0.009=6.977\lt7$$
Albert said:now ,try to prove it,without using calculator
you don't have to compute the square roots of those numbers ,use some tricksMarkFL said:It is fairly easy to compute square roots by hand to two decimal places...[cs=Spiteful]spiteful.gif[/cs]
very smart!anemone said:I am going to use a trick someone taught me from this forum to solve for this challenge:
Note that
$288<289\,\,\implies2(12^2)<17^2$ or $(\sqrt{2}<\dfrac{17}{12})---(1)$
$80<81\,\,\implies5(4^2)<9^2$ or $(\sqrt{5}<\dfrac{9}{4})---(2)$
$99<100\,\,\implies11(3^2)<10^2$ or $(\sqrt{11}<\dfrac{10}{3})---(3)$
Adding the inequalities in (1), (2) and (3) up gives us the answer:
$\sqrt{2}+\sqrt{5}+\sqrt{11}<\dfrac{17}{12}+\dfrac{9}{4}+\dfrac{10}{3}=7$