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What is the ratio of the length of the diagonal of a perfect square to an edge?. What is the ratio of the length of the diagonal of a perfect square to an edge?. What is the ratio of the length of the diagonal of a perfect square to an edge?. The white area in the top square is ( a 2 )/2.
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What is the ratio of the length of the diagonal of a perfect square to an edge?
What is the ratio of the length of the diagonal of a perfect square to an edge?
What is the ratio of the length of the diagonal of a perfect square to an edge? The white area in the top square is (a2)/2.
What is the ratio of the length of the diagonal of a perfect square to an edge? The white area in the top square is (a2)/2. So the white area in the lower square is 2a2.
What is the ratio of the length of the diagonal of a perfect square to an edge? The white area in the top square is (a2)/2. So the white area in the lower square is 2a2. But this area can also be expressed as b2.
What is the ratio of the length of the diagonal of a perfect square to an edge? The white area in the top square is (a2)/2. So the white area in the lower square is 2a2. But this area can also be expressed as b2. Thus, b2 = 2a2.
What is the ratio of the length of the diagonal of a perfect square to an edge? The white area in the top square is (a2)/2. So the white area in the lower square is 2a2. But this area can also be expressed as b2. Thus, b2 = 2a2. Or, (b/a)2 = 2.
We conclude that the ratio of the diagonal to the edge of a square is the square root of 2, which can be written as √2 or 21/2.
So √2 is with us whenever a perfect square is. • For a period of time, the ancient Greek mathematicians believed any two distances are commensurate (can be co-measured).
So √2 is with us whenever a perfect square is. • For a period of time, the ancient Greek mathematicians believed any two distances are commensurate (can be co-measured). • For a perfect square this means a unit of measurement can be found so that the side and diagonal of the square are both integer multiples of the unit.
This means √2 would be the ratio of two integers. • A ratio of two integers is called a rational number.
This means √2 would be the ratio of two integers. • A ratio of two integers is called a rational number. • To their great surprise, the Greeks discovered √2 is not rational.
This means √2 would be the ratio of two integers. • A ratio of two integers is called a rational number. • To their great surprise, the Greeks discovered √2 is not rational. • Real numbers that are not rational are now called irrational.
This means √2 would be the ratio of two integers. • A ratio of two integers is called a rational number. • To their great surprise, the Greeks discovered √2 is not rational. • Real numbers that are not rational are now called irrational. • We believe √2 was the very first number known to be irrational. This discovery forced a rethinking of what “number” means.
We will present a proof that √2 is not rational. • Proving a negative statement usually must be done by assuming the logical opposite and arriving at a contradictory conclusion.
We will present a proof that √2 is not rational. • Proving a negative statement usually must be done by assuming the logical opposite and arriving at a contradictory conclusion. • Such an argument is called a proof by contradiction.
Theorem: There is no rational number whose square is 2. Proof : Assume, to the contrary, that √2is rational.
Theorem: There is no rational number whose square is 2. Proof : Assume, to the contrary, that √2is rational. So we can write√2= n/m with n and m positive integers.
Theorem: There is no rational number whose square is 2. Proof : Assume, to the contrary, that √2is rational. So we can write√2= n/m with n and m positive integers. Among all the fractions representing √2, we select the one with smallest denominator.
So if √2 is rational (√2= n/m) then an isosceles right triangle with legs of length m will have hypotenuse of length n= √2m. n = √2m
So if √2 is rational (√2= n/m) then an isosceles right triangle with legs of length m will have hypotenuse of length n= √2m. Moreover, for a fixed unit, we can take ΔABC to be the smallest isosceles right triangle with integer length sides. n = √2m
Now, for the basic trick. Bisect the angle at A and fold the edge AB along the edge AC.
Now, for the basic trick. Bisect the angle at A and fold the edge AB along the edge AC.
Now, for the basic trick. Bisect the angle at A and fold the edge AB along the edge AC. This creates a new triangle ΔDEC with the angle at E being a right angle and the angle at C still being 45⁰.
Now, for the basic trick. Bisect the angle at A and fold the edge AB along the edge AC. This creates a new triangle ΔDEC with the angle at E being a right angle and the angle at C still being 45⁰. AE=AB=m
Now, for the basic trick. Bisect the angle at A and fold the edge AB along the edge AC. This creates a new triangle ΔDEC with the angle at E being a right angle and the angle at C still being 45⁰. AE=AB=n EC=AC-AE
Now, for the basic trick. Bisect the angle at A and fold the edge AB along the edge AC. This creates a new triangle ΔDEC with the angle at E being a right angle and the angle at C still being 45⁰. AE=AB=n EC=AC-AE=n-m
Now, for the basic trick. Bisect the angle at A and fold the edge AB along the edge AC. This creates a new triangle ΔDEC with the angle at E being a right angle and the angle at C still being 45⁰. AE=AB=n EC=AC-AE=n-m BD=DE
Now, for the basic trick. Bisect the angle at A and fold the edge AB along the edge AC. This creates a new triangle ΔDEC with the angle at E being a right angle and the angle at C still being 45⁰. AE=AB=n EC=AC-AE=n-m BD=DE=EC=n-m
But, if BD=DE=EC=n-m and BC=m,
But, if BD=DE=EC=n-m and BC=m, then DC=BC-BD
But, if BD=DE=EC=n-m and BC=m, then DC=BC-BD=m-(n-m)
But, if BD=DE=EC=n-m and BC=m, then DC=BC-BD=m-(n-m)=2m-n.
But, if BD=DE=EC=n-m and BC=m, then DC=BC-BD=m-(n-m)=2m-n.
But, if BD=DE=EC=n-m and BC=m, then DC=BC-BD=m-(n-m)=2m-n. Since n and m are integers, n-m and 2m-n are integers and ΔDEC is an isosceles right triangle with integer side lengths smaller than ΔABC .
This contradicts our choice of ΔABC as the smallest isosceles right triangle with integer side lengths for a given fixed unit of length.
This contradicts our choice of ΔABC as the smallest isosceles right triangle with integer side lengths for a given fixed unit of length. This means our assumption that √2is rational is false. Thus there is no rational number whose square is 2. QED
This beautiful proof was adapted from Tom Apostol: “Irrationality of the Square Root of Two: A Geometric Proof”, American Mathematical Monthly,107, 841-842 (2000).
This beautiful proof was adapted from Tom Apostol: “Irrationality of the Square Root of Two: A Geometric Proof”, American Mathematical Monthly,107, 841-842 (2000). Behold, √2is irrational!