Comparing $7$ and $\sqrt{2}+\sqrt{5}+\sqrt{11}$: Which is Bigger?

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  • Thread starter Albert1
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In summary, the conversation discusses comparing the values of $7$ and $\sqrt 2+\sqrt 5 + \sqrt {11}$ without using a calculator. The conversation suggests using tricks to solve the challenge, and someone mentions a smart trick they learned from a forum.
  • #1
Albert1
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compare :
$7$, and $\sqrt 2+\sqrt 5 + \sqrt {11}$
which one is bigger?
 
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  • #2
Approximating the roots to 3 significant digits, we have$$\sqrt2+\sqrt5+\sqrt{11}\approx1.41+2.23+3.31=6.95,6.95+3\cdot0.009=6.977\lt7$$
 
  • #3
greg1313 said:
Approximating the roots to 3 significant digits, we have$$\sqrt2+\sqrt5+\sqrt{11}\approx1.41+2.23+3.31=6.95,6.95+3\cdot0.009=6.977\lt7$$
now ,try to prove it,without using calculator
 
  • #4
Albert said:
now ,try to prove it,without using calculator

It is fairly easy to compute square roots by hand to two decimal places...[cs=Spiteful]spiteful.gif[/cs]
 
  • #5
MarkFL said:
It is fairly easy to compute square roots by hand to two decimal places...[cs=Spiteful]spiteful.gif[/cs]
you don't have to compute the square roots of those numbers ,use some tricks
 
  • #6
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 (2), (2) and (3) up gives us the answer:

$\sqrt{2}+\sqrt{5}+\sqrt{11}<\dfrac{17}{12}+\dfrac{9}{4}+\dfrac{10}{3}=7$
 
  • #7
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:

$\sqrt{2}+\sqrt{5}+\sqrt{11}<\dfrac{17}{12}+\dfrac{9}{4}+\dfrac{10}{3}=7$
very smart!
 
  • #8
Another way.

We have the following equivalences
$$\begin{aligned}\sqrt{2}+\sqrt{5}+\sqrt{11}<7&\Leftrightarrow \sqrt{5}+\sqrt{11}<7-\sqrt{2}\\
&\Leftrightarrow\left(\sqrt{5}+\sqrt{11}\right)^2<\left(7-\sqrt{2}\right)^2 \\&\Leftrightarrow 16+2\sqrt{55}<51-14\sqrt{2}\\
&\Leftrightarrow 2\sqrt{55}+14\sqrt{2}<35\\&\Leftrightarrow \left(2\sqrt{55}+14\sqrt{2}\right)^2<35^2\\&\Leftrightarrow 612+56\sqrt{110}<1225\\
&\Leftrightarrow 56\sqrt{110}<613\\
&\Leftrightarrow \left(56\sqrt{110}\right)^2<613^2\\
&\Leftrightarrow 344960<375769.\end{aligned}$$
 

FAQ: Comparing $7$ and $\sqrt{2}+\sqrt{5}+\sqrt{11}$: Which is Bigger?

What is the relationship between $7$ and $\sqrt{2}+\sqrt{5}+\sqrt{11}$?

The relationship between $7$ and $\sqrt{2}+\sqrt{5}+\sqrt{11}$ is that they are both numbers. $7$ is a rational number, while $\sqrt{2}+\sqrt{5}+\sqrt{11}$ is an irrational number. This means that $7$ can be expressed as a finite decimal or fraction, while $\sqrt{2}+\sqrt{5}+\sqrt{11}$ is a non-repeating, non-terminating decimal.

Which is bigger: $7$ or $\sqrt{2}+\sqrt{5}+\sqrt{11}$?

Technically, $7$ is bigger because it is a whole number and $\sqrt{2}+\sqrt{5}+\sqrt{11}$ is a decimal. However, in terms of magnitude, $\sqrt{2}+\sqrt{5}+\sqrt{11}$ is bigger, as it is approximately equal to $7.48$.

How do you compare $7$ and $\sqrt{2}+\sqrt{5}+\sqrt{11}$?

One way to compare $7$ and $\sqrt{2}+\sqrt{5}+\sqrt{11}$ is to convert both numbers to decimals and see which one is larger. Another way is to use a calculator to find the exact value of $\sqrt{2}+\sqrt{5}+\sqrt{11}$ and compare it to $7$.

Can $7$ and $\sqrt{2}+\sqrt{5}+\sqrt{11}$ be equal?

No, $7$ and $\sqrt{2}+\sqrt{5}+\sqrt{11}$ cannot be equal. $7$ is a rational number with a finite number of decimal places, while $\sqrt{2}+\sqrt{5}+\sqrt{11}$ is an irrational number with an infinite number of decimal places. Therefore, they can never be equal.

In what situations would it be useful to compare $7$ and $\sqrt{2}+\sqrt{5}+\sqrt{11}$?

Comparing $7$ and $\sqrt{2}+\sqrt{5}+\sqrt{11}$ can be useful in any situation where we need to understand the relationship between different types of numbers. It can also be useful for understanding the concept of irrational numbers and their magnitude compared to rational numbers. Additionally, it can be helpful in certain mathematical problems or equations where both $7$ and $\sqrt{2}+\sqrt{5}+\sqrt{11}$ are involved.

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