Conundrums

Klaas van Aarsen

MHB Seeker
Staff member
Let me start this thread for math conundrums.
That is, proofs that we know cannot be right, but where it's not immediately (too) obvious what the problem is.
If you have one of your own that you feel like sharing, please post it!

I'll start:
\begin{array}{rcl}
x&=&(\pi+3)/2 \\
2x&=&\pi+3 \\
2x(\pi-3)&=&(\pi+3)(\pi-3) \\
2\pi x - 6x &=& \pi^2 -9 \\
9-6x &=& \pi^2-2\pi x \\
9-6x+x^2 &=& \pi^2-2\pi x + x^2 \\
(3-x)^2 &=& (\pi-x)^2 \\
3-x &=& \pi -x \\
\pi &=& 3
\end{array}

Please feel free to share one of your own!

Jameson

Administrator
Staff member
Not a math proof per say, and I admit I found this on Stack Exchange, but one could argue this is a sort of geometric proof.

Below is the never ending chocolate bar! Where is the error?

RLBrown

Active member
I thought it might be fun to ask is 1=2?
I will try 3 ways to convince you.

ALGEBRA:
--------
x = y
x^2 = xy //////////////// Mult by x
2x^2 = x^2 + xy ///////// Add x^2
2x^2 - 2xy = x^2 - 2xy // Subt 2xy
2x(x-y) = x(x-y) //////// Factor
2x = x ////////////////// Cancel x-y
2 = 1 /////////////////// Divide x

CALCULUS:
---------
2X = X+X
3X = X+X+X
NX = X+X+X+ ... +X
.....|__N times__|

XX = X+X+X+ ... +X
.....|__X times__|

(X^2) = (X+X+X+ ... +X)
.........|__X times__|

2X = 1+1+1+ ... +1 :derivative of both sides
.....|__X times__|

2X = X
2 = 1

Physics (electronics)
---------------------
For a capacitor
E = (1/2) C*V^2
Q = CV
where
E = energy in capacitor (joules)
V = voltage across capacitor (volts)
Q = charge in capacitor (coulombs)
C = capacitance (in farads)

CASE 1: (initial condition)
I have 2 capacitors C1 and C2 both 1 farad
V1=0
V2=2
Total Energy = E1 + E2 = 2 joules

CASE 2: (final condition)
I will connect the two capacitors across
each other (the charge will redistribute
half in each)
V1 = V2 = 1
No charge was lost.
Physics says that total energy is conserved
so Total Energy = 2 joules (as before)
OPPS! the formula says
Total Energy = 2 * (1/2)(1)(1^2) = 1 joule

Since the total energy is conserved, 2=1
=============================================================

Evgeny.Makarov

Well-known member
MHB Math Scholar
Theorem. All positive integers are equal.

Proof. It suffices to show that for any two positive integers, $A$ and $B$, $A = B$. Further, it suffices to show that for all $N > 0$, if $A$ and $B$ are positive integers which satisfy $\max(A, B) = N$ then $A = B$.

We proceed by induction. If $N = 1$, then $A$ and $B$, being positive integers, must both be 1. So $A = B$.

Assume that the theorem is true for some value $k$. Take $A$ and $B$ with $\max(A, B) = k+1$. Then $\max((A-1), (B-1)) = k$. And hence $(A-1) = (B-1)$. Consequently, $A = B$.

Hence, $A = B$ for all positive integers $A$ and $B$ by induction. Q.E.D.