Proving $\sqrt{2}+\sqrt{3} \gt \pi$: A Mathematical Challenge

In summary, the mathematical challenge of proving √2 + √3 > π is to show that the sum of two irrational numbers, √2 and √3, is greater than the irrational number π. This statement is true for all values of √2 and √3, and can be proven using techniques such as real analysis, the properties of irrational numbers, and inequalities and limits. This statement is important in mathematics as it demonstrates the complexity and relationship of irrational numbers and has practical applications in fields such as physics, engineering, and finance. A real-life example that illustrates this statement is the construction of a regular pentagon using only a ruler and compass.
  • #1
anemone
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Prove $\sqrt{2}+\sqrt{3}\gt \pi$.
 
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  • #2
anemone said:
Prove $\sqrt{2}+\sqrt{3}\gt \pi$.
$\begin{vmatrix}
\sqrt 3-\dfrac{\pi}{2}
\end{vmatrix}>\begin{vmatrix}
\sqrt 2-\dfrac{\pi}{2}
\end{vmatrix}=\begin{vmatrix}
\dfrac{\pi}{2}-\sqrt 2
\end{vmatrix}$
we have:
$\sqrt 3-\dfrac{\pi}{2}>\dfrac {\pi}{2}-\sqrt 2>0$
$\therefore \sqrt 3+\sqrt 2>\pi$
the proof is done
 
  • #3
Albert said:
$\begin{vmatrix}
\sqrt 3-\dfrac{\pi}{2}
\end{vmatrix}>\begin{vmatrix}
\sqrt 2-\dfrac{\pi}{2}
\end{vmatrix}=\begin{vmatrix}
\dfrac{\pi}{2}-\sqrt 2
\end{vmatrix}$
we have:
$\sqrt 3-\dfrac{\pi}{2}>\dfrac {\pi}{2}-\sqrt 2>0$
$\therefore \sqrt 3+\sqrt 2>\pi$
the proof is done

Very well done, Albert!(Clapping)
 
  • #4
Albert said:
$\begin{vmatrix}
\sqrt 3-\dfrac{\pi}{2}
\end{vmatrix}>\begin{vmatrix}
\sqrt 2-\dfrac{\pi}{2}
\end{vmatrix}=\begin{vmatrix}
\dfrac{\pi}{2}-\sqrt 2
\end{vmatrix}$
we have:
$\sqrt 3-\dfrac{\pi}{2}>\dfrac {\pi}{2}-\sqrt 2>0$
$\therefore \sqrt 3+\sqrt 2>\pi$
the proof is done
[sp]I don't understand this argument. Why is $\Bigl|\sqrt 3-\dfrac{\pi}{2}\Bigr| > \Bigl|\dfrac{\pi}{2}-\sqrt 2\Bigr|$?[/sp]
 
  • #5
Opalg said:
[sp]I don't understand this argument. Why is $\Bigl|\sqrt 3-\dfrac{\pi}{2}\Bigr| > \Bigl|\dfrac{\pi}{2}-\sqrt 2\Bigr|$?[/sp]

if $\Bigl|A\Bigr| > \Bigl|B\Bigr|$
and $\Bigl|B\Bigr| = \Bigl|C\Bigr|$
then
$\Bigl|A\Bigr| > \Bigl|C\Bigr|$
 
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  • #6
Opalg said:
[sp]I don't understand this argument. Why is $\Bigl|\sqrt 3-\dfrac{\pi}{2}\Bigr| > \Bigl|\dfrac{\pi}{2}-\sqrt 2\Bigr|$?[/sp]

Albert said:
if $\Bigl|A\Bigr| > \Bigl|B\Bigr|$
and $\Bigl|B\Bigr| = \Bigl|C\Bigr|$
then
$\Bigl|A\Bigr| > \Bigl|C\Bigr|$

I have long thought about the validity of Albert's solution, but his argument I think is also built on the facts that he knows $\sqrt{3}>\dfrac{\pi}{2}$ and also, $\sqrt{2}<\dfrac{\pi}{2}$.

Anyway, here is my solution:

We could use the well-known inequality that says $\dfrac{22}{7}>\pi$ to assist in my method of proving, as shown below:

Note that

$7938>7921$

$2(63)^2>89^2$

$\sqrt{2}>\dfrac{89}{63}$ and

$762048>760384$

$3(504)^2>872^2$

$\sqrt{3}>\dfrac{872}{504}$,

$\therefore \sqrt{2}+\sqrt{3}>\dfrac{22}{7}>\pi$ and we're hence done.
 
  • #7
it is very easy to prove $\sqrt 3>\dfrac {\pi}{2}$ , and $\sqrt 2<\dfrac {\pi}{2}$
for :
$2<2.25=\dfrac{3^2}{4}<(\dfrac {\pi}{2})^2<\dfrac {3.2^2}{4}=2.56<3$
 
Last edited by a moderator:

FAQ: Proving $\sqrt{2}+\sqrt{3} \gt \pi$: A Mathematical Challenge

What is the mathematical challenge of proving √2 + √3 > π?

The mathematical challenge of proving √2 + √3 > π is to show that the sum of the irrational numbers √2 and √3 is greater than the irrational number π. This challenge requires the use of advanced mathematical techniques and concepts, such as real analysis and the properties of irrational numbers.

Is this statement true for all values of √2 and √3?

Yes, this statement is true for all values of √2 and √3. The inequality √2 + √3 > π holds for any values of √2 and √3, as long as they are both irrational numbers.

What are some techniques that can be used to prove this statement?

Some techniques that can be used to prove √2 + √3 > π include real analysis, the properties of irrational numbers, and the use of inequalities and limits. Other advanced mathematical concepts, such as calculus and trigonometry, may also be used in the proof.

Why is this statement important in mathematics?

This statement is important in mathematics because it demonstrates the relationship between different types of irrational numbers. It also highlights the complexity of irrational numbers and the need for advanced mathematical techniques to understand and prove their properties. Additionally, this statement has applications in various fields such as physics, engineering, and finance.

Are there any real-life examples that illustrate this statement?

One real-life example that illustrates this statement is the construction of a regular pentagon using only a ruler and compass. The length of one side of a regular pentagon is √5, which can be expressed as √2 + √3. This shows that the sum of two irrational numbers, √2 and √3, is greater than the irrational number π, which is the ratio of the circumference of a circle to its diameter.

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