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Proof that all integers are equal June 1, 2017

Posted by Ezra Resnick in Math.
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In particular, we prove that n = -1 for any integer n.

First, assume that n is odd. That means that n – 1 is even, so n – 1 = 2k for some integer k.

The following identity is trivially true for any n:

n2 – 1 = (n – 1)(n + 1)

We can substitute 2k for (n – 1):

n2 – 1 = 2k(n + 1) = 2kn + 2k

Next, we subtract (2knn – 1) from both sides of the equation, yielding:

n2 – 2kn – n = 2k – n + 1

Factoring out (n – 2k – 1) produces:

n(n – 2k – 1) = –(n – 2k – 1)

Now we simply divide both sides of the equation by (n – 2k – 1), and voila:

n = -1

Alternatively, assume that n is even. That means that n – 1 is odd, so n – 1 = 2k + 1 for some integer k.

Once again, we begin with the identity:

n2 – 1 = (n – 1)(n + 1)

This time we substitute (2k + 1) for (n – 1):

n2 – 1 = (2k + 1)(n + 1) = 2kn + 2k + n + 1

We subtract (2kn + 2n – 1) from both sides of the equation:

n2 – 2kn – 2n = 2k – n + 2

We factor out (n – 2k – 2):

n(n – 2k – 2) = –(n – 2k – 2)

And now we simply divide both sides of the equation by (n – 2k – 2) to complete the proof:

n = -1

Q.E.D.

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Comments»

1. Ubi Dubium - June 1, 2017

Ah. I see what you did there. 😛


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