0 = 1: A Myriad of Proofs
Or, How My PhD became Useless Overnight
Many of these “proofs” come from the “Fallacies, Flaws, and Flimflam” section of The College
Mathematics Journal. If you know of, or find some others, let me know at vestal@mwsc.edu.
Theorem Let x and y denote distinct real numbers. Then
0  1  x  y.
PROOF. Straight-forward approach:
0  1  0  x  y   1 x  y 
 0 x y
 x  y.
Contradiction approach:
 Suppose that 0  1 and x  y. This is a contradiction (0 = 1), therefore x  y.
 
Suppose that x  y and 0  1. This is a contradiction (x = y), therefore 0  1.
Note that in both cases, the contradiction in the hypothesis was all that mattered. In general, a
contradiction can prove any statement to be true.
An algebraic proof
x  1  x2  x
 x2  1  x  1
1111
 1 1  1 1
  x  1 x  1  x  1
 1  11  1  1  1
  x  1  x  1  x  1
  0  2  0
 x 1  1
 2 1
x0
1 0
An induction proof
Theorem. Let x and y be positive integers. Then x  y.
PROOF. We proceed by induction on n  max x, y.
Basis Step: n  1.
If x and y are positive integers whose maximum is 1, then clearly x  y.
Induction Step: n  n  1
Suppose that for any two positive integers x and y, if max x, y  n, then x  y. Now suppose
that x and y are positive integers with max x, y  n  1. Then max x 1, y 1  n, which, by
the induction hypothesis implies that x  1  y  1, and hence x  y.
A Limit proof involving Slant Asymptotes
Find the slant asymptote of the rational function
y
x2  2 x  2
.
x 1
Using long division, we get
x2  2 x  2 x2  2 x  1  1 x2  2 x  1
1
1



 x 1
,
x 1
x 1
x 1
x 1
x 1
and thus the oblique asymptote is y  x  1.
On the other hand, dividing by x yields
x2  2x  2 x  2  2 / x

,
x 1
1  1/ x
and letting x  , this becomes
x2
,
1
so that the oblique asymptote is y  x  2. This implies that 1  2.
A Calculus proof involving derivatives
Since
x2  x  x 
 x,
x times
it follows that

d 2
d 
 x    x  x   x  ,
dx
dx 
x times

i.e. 2 x  1  1 
 1  x.
x times
Since this is true for all x, it is true for x  1, and therefore 2  1.
A Calculus proof involving integration
1
Consider  dx. Using integration by parts with
x
1
u
dv  dx
x
1
du  2 dx v  x
x
yields
1
x
1
 x dx  1   x 2 dx  1   x dx
and thus 0  1.
A Calculus proof involving definite integrals

We evaluate  sin 2 x dx in two ways.
0

1:

1  cos 2 x
x cos 2 x
dx  
2
2
4
0
2
 sin x dx  
0
x 

x 0

2
Using the substitution u  sin x, we get du  cos x dx  1  sin 2 x dx,
du
du
dx 

. Thus the integral becomes
1  sin 2 x
1 u2

0
u 2 du
2
sin
x
dx

0
0 1  u 2  0.

Therefore  0.
2
2:
so that
A Calculus proof involving Series
Using the Taylor series for the natural logarithm
x 2 x3 x 4
ln 1  x   x     ,
2 3 4
with x  1, we get
1 1 1 1 1 1 1 1 1 1 1
ln 2  1            
2 3 4 5 6 7 8 9 10 11 12
1 1 1 1 1 1 1 1 1 1
 1          
2 4 3 6 8 5 10 12 7 14
 1 1 1 1 1 1 1  1 1 1 
 1                  
 2  4  3 6  8  5 10  12  7 14 
1 1 1 1 1 1 1
       
2 4 6 8 10 12 14
1 1 1 1 1 1 1 1 1 1 1 1

 1             
2  2 3 4 5 6 7 8 9 10 11 12

1
 ln 2.
2
The only number equal to half of itself is zero, thus ln 2  0.
A Linear Algebra proof
Choose a positive integer n and let S be the set of all real solutions  x1 , x2 , , xn , xn1  to the
equation
x1  x2   xn  xn1  1.
(*)
Since any choice of x1 , x2 , , xn uniquely determines the value of xn 1 , the mapping
f : Rn1  Rn defined by f  x1 , x2 ,
, xn , xn1    x1 , x2 ,
, xn  maps S one-to-one onto R n .
Moreover, f preserves addition and scalar multiplication, so that S is in fact isomorphic to R n .
Therefore, S must be a vector space, and thus contains the zero vector  0,0, ,0  . But if this
zero vector satisfies (*), then 0 must equal 1.
A Complex Analysis proof
Picard’s Theorem: If f  z  is a nonconstant, entire function (differentiable everywhere), then f
takes on every complex value with the possible exception of one complex number.
For example, the function f  z   e z takes on every complex value except for 0.
Theorem. 0  1
z
PROOF. Applying Picard’s Theorem to the nonconstant, analytic function f  z   ee , this
function takes on all complex values, with at most one exception.
z
ee  0 : This is impossible, because e to any power can never be zero.
ee  1: This is impossible, because ee  1  e z  0.
Therefore, 0  1.
z
z
A Physics Proof (with some trigonometry)
Suppose that we have a stick one foot long leaning against a wall. Find the angle  which will
maximize the value of x 2  y. Using x  cos and y  sin  , we maximize cos 2   sin  to
obtain    / 6. Repeating the problem using 12 inches, we have x  12cos and y  12sin  ,
and the maximum of 144 cos 2   12sin  occurs when sin   1/ 24, which is not  / 6.
Therefore, one foot does not equal 12 inches.
1 foot =
12 inches
y

x
A Proof involving exponents
x
 2.
x
 2, it follows that x 2  2, and thus x  2. On the other hand, consider the equation
Solve x x
From x x
yy
y
 4. As before, we get
yy
This gives us
2 2
2
2
 4.
y
4 
y4  4 
y  2.