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rational/rational=irrational=π How?


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Hello everyone,


I'm new here and I have a question;

So we know that π is a circle's circumference/diameter

So 2 rational numbers dividing and cames out as irrational.

Can you explain me how is this happening and is there any examples like that?

Sorry for bad English btw.


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Hi Eren, welcome to the forum.


It is NOT two rational numbers being divided. That's actually the definition of an irrational number, there are no two numbers that we can use in a ratio or division to make that irrational number.




However, a good example of two rational and measurable numbers 'making' an irrational one is a right triangle with the two sides next to the right angle having length one. The hypotenuse of that triangle will have length [math]\sqrt{2}[/math] which is also an irrational number.

Edited by Bignose
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You are falling into a mental trap by thinking of circumference/diameter as two rational numbers. Consider this: If the diameter is 1, the circumference is 1 times pi, which is irrational. If the diameter is 2, the circumference is 2 times pi-- which is still irrational. In fact, in any circle, if you could measure it to an infinite number of decimal places, you would discover that one of those two measurements is an irrational number. Dividing two numbers, one of which is irrational, gives you an irrational result.

Edited by OldChemE
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If fact, what your argument proves is: "if the radius of a circle is a rational number the circumference can't be a rational number" and "if the circumference of a circle is a rational number then the radius can't be a rational number."

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Correct - the quotient of any two rational numbers is always a rational number. Say we have r1 = a/b and r2 = c/d, where a..d are integers. That's the definition of a rational number (quotient of two integers). So now


r1/r2 = (a/b) / (c/d) = (a/b) * (d/c) = (ad)/(bc).


The products ad and bc are both the products of integers, and thus are integers. The quotient of those two products is thus a rational number.


The previous answers point out the misstep in your original reasoning.

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