Can Two Numbers from a Set of Thirteen Satisfy This Mathematical Inequality?

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In summary, it is possible to choose two distinct numbers, $a$ and $b$, from a set of thirteen real numbers such that $0<\dfrac{a-b}{1+ab}<2-\sqrt{3}$. This solution requires the chosen numbers to be specific, not just any two numbers from the set.
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anemone
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Show that among any thirteen real and distinct numbers, it is possible to choose two, says, $a$ and $b$ such that $0<\dfrac{a-b}{1+ab}<2-\sqrt{3}$.
 
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anemone said:
Show that among any thirteen real and distinct numbers, it is possible to choose any two, says, $a$ and $b$ such that $0<\dfrac{a-b}{1+ab}<2-\sqrt{3}$.

It cannot any 2 because chosen a and b have to be specific here , Here is the solution

write the numbers has $\tan\ x$. Now x shall be between $-90^0$ to $90^0$. divide -90 to 90 degrees of 12 intervals of 15 degrees each. Now as there are 13 numbers 2 numbers have to in one interval.
say the angles are x, y and tan x = a , tan y = b and a > b
$\dfrac{a-b}{1+ab}= \tan( x - y) \lt tan \ 15 \lt 2-\sqrt{3}$
 
  • #3
kaliprasad said:
It cannot any 2 because chosen a and b have to be specific here,

Hey kaliprasad, I thought I've already replied to this challenge days ago, so... sorry for the late reply!:eek:

Yes, you're right, the wording of the problem sounds wrong, I'll change it so that it becomes a valid and sound problem, thanks for pointing this out...:)

Here is the solution
kaliprasad said:
write the numbers has $\tan\ x$. Now x shall be between $-90^0$ to $90^0$. divide -90 to 90 degrees of 12 intervals of 15 degrees each. Now as there are 13 numbers 2 numbers have to in one interval.
say the angles are x, y and tan x = a , tan y = b and a > b
$\dfrac{a-b}{1+ab}= \tan( x - y) \lt tan \ 15 \lt 2-\sqrt{3}$

Needless to say, you're another math prodigy at our site! Thanks for your neat proof and thanks for participating, also, glad to see you around at the challenge problems sub forum, kali!
 

FAQ: Can Two Numbers from a Set of Thirteen Satisfy This Mathematical Inequality?

1. What does the inequality 0<(a-b)/(1+ab)<2-√3 represent?

The inequality represents a certain range of values that the expression (a-b)/(1+ab) can take. The expression is bounded by 0 on the lower end and 2-√3 on the upper end.

2. How do you prove that 0<(a-b)/(1+ab)<2-√3 is true?

This inequality can be proven by breaking it down into two separate inequalities: 0<(a-b)/(1+ab) and (a-b)/(1+ab)<2-√3. Each inequality can then be solved algebraically and the solutions can be combined to show that the original inequality is true.

3. What are the restrictions on the variables a and b for the inequality to hold?

The variables a and b must be real numbers, and a cannot equal b. Additionally, the expression (1+ab) cannot equal 0, so a and b cannot be reciprocals of each other.

4. Can the inequality 0<(a-b)/(1+ab)<2-√3 be simplified?

Yes, the inequality can be simplified by rationalizing the denominator of the expression (a-b)/(1+ab). This results in the inequality 0<(a^2-ab+b)/(1+ab)^2<2-√3. However, this does not change the range of values that the expression can take.

5. In what scenarios would the inequality 0<(a-b)/(1+ab)<2-√3 be useful in scientific research?

This inequality can be useful in situations where researchers are trying to find values that fall within a certain range. For example, it could be used in analyzing data from experiments or simulations, or in setting boundaries for mathematical models. It could also be used in optimization problems where the goal is to find the optimal values for a and b within the given constraints.

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