Download Fractions don`t exist

Fractions don’t exist
Theorem
The smallest positive number is 1.
()
August 15, 2014
1/5
Fractions don’t exist
Theorem
The smallest positive number is 1.
Proof.
Let x be the smallest positive number.Clearly x ≤ 1.
()
August 15, 2014
1/5
Fractions don’t exist
Theorem
The smallest positive number is 1.
Proof.
Let x be the smallest positive number.Clearly x ≤ 1.
Now x 2 is also positive, so by minimality of x, x ≤ x 2 .
()
August 15, 2014
1/5
Fractions don’t exist
Theorem
The smallest positive number is 1.
Proof.
Let x be the smallest positive number.Clearly x ≤ 1.
Now x 2 is also positive, so by minimality of x, x ≤ x 2 .
Divide both sides by (the positive number) x to get 1 ≤ x. Thus 1 ≤ x ≤ 1.
Hence x = 1.
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August 15, 2014
1/5
All numbers are equal
Theorem
All numbers are equal
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August 15, 2014
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All numbers are equal
Theorem
All numbers are equal
Proof.
Pick any two numbers, m and n say.
()
August 15, 2014
2/5
All numbers are equal
Theorem
All numbers are equal
Proof.
Pick any two numbers, m and n say.
mx +ny
Consider T = x +y , which we can calculate for various values of x and y.
()
August 15, 2014
2/5
All numbers are equal
Theorem
All numbers are equal
Proof.
Pick any two numbers, m and n say.
mx +ny
Consider T = x +y , which we can calculate for various values of x and y.
ny
Firstly, whenever x = 0, we have T = y = n.
On the other hand if y = 0, then T = mx
x = m.
()
August 15, 2014
2/5
All numbers are equal
Theorem
All numbers are equal
Proof.
Pick any two numbers, m and n say.
mx +ny
Consider T = x +y , which we can calculate for various values of x and y.
ny
Firstly, whenever x = 0, we have T = y = n.
On the other hand if y = 0, then T = mx
x = m.
So if x = 0, T = n and if y = 0, T = m. Setting both x and y = 0 gives
n = T = m. Hence m = n.
()
August 15, 2014
2/5
All numbers are equal
Theorem
All numbers are equal
Proof.
Pick any two numbers, m and n say.
mx +ny
Consider T = x +y , which we can calculate for various values of x and y.
ny
Firstly, whenever x = 0, we have T = y = n.
On the other hand if y = 0, then T = mx
x = m.
So if x = 0, T = n and if y = 0, T = m. Setting both x and y = 0 gives
n = T = m. Hence m = n.
This holds for any m, n so all numbers are equal.
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August 15, 2014
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Contradiction Proofs
Theorem
There is no point on the circumference of a circle nearest to a given point
inside it.
P
r
t
O
a
C
Choose a point inside the circle and set it to be the origin O of Cartesian
coordinates, with the x-axis joining O to the centre C. Let the coordinates
of C be (a , 0) and the radius be r, so (x − a )2 + y 2 = r 2 or
x 2 + y 2 − 2ax + a 2 − r 2 = 0.
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August 15, 2014
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The distance from O to the point P (x , y ) of the circle is t, where
t2 = x2 + y2
so, since P is on the circle
t 2 = 2ax − a 2 + r 2 .
The distance OP is least, or greatest, at points for which
differentiating,
dt
2t dx
= 2a .
dt
dx
= 0. But,
dt
So dx
can only be 0 when a = 0, that is, if O is the centre of the circle, in
which case all points on the circumference are equidistant from O. When
a , 0 then there is no point on the circumference whose distance from O
is either a maximum or a minimum.
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August 15, 2014
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Corollary
The only point inside a circle is its centre
Proof.
Suppose this is false, so let a be a point in the circle, not the centre. Then
there is no point on the circumference nearest to a. This is absurd, so we
have a contradiction. Therefore the only point inside a circle is its
centre.
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August 15, 2014
5/5