**n equals n plus one**

Proof:

(n+1)^2 = n^2 + 2*n + 1

Bring 2n+1 to the left:

(n+1)^2 - (2n+1) = n^2

Substract n(2n+1) from both sides and factoring, we have:

(n+1)^2 - (n+1)(2n+1) = n^2 - n(2n+1)

Adding 1/4(2n+1)^2 to both sides yields:

(n+1)^2 - (n+1)(2n+1) + 1/4(2n+1)^2 = n^2 - n(2n+1) + 1/4(2n+1)^2

This may be written:

[ (n+1) - 1/2(2n+1) ]^2 = [ n - 1/2(2n+1) ]^2

Taking the square roots of both sides:

(n+1) - 1/2(2n+1) = n - 1/2(2n+1)

Add 1/2(2n+1) to both sides:

n+1 = n

**one is negative one**

Proof:

1 = sqrt(1) = sqrt(-1 * -1) = sqrt(-1) * sqrt(-1) = 1^ = -1

Also one can disprove the axiom that things equal to the same thing are equal to each other.

1 = sqrt(1)

-1 = sqrt(1)

Therefore 1 = -1

**As an alternative method for solving:**

Theorem: 1 = -1

Proof:

x=1

x^2=x

x^2-1=x-1

(x+1)(x-1)=(x-1)

(x+1)=(x-1)/(x-1)

x+1=1

x=0

0=1

=> 0/0=1/1=1

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