Top 10 Best Examples Of Illegal Math To Fool People- The Top Ten Today

Mathematics with a tincture of intelligence and cleverness even if mixed with some BS will confuse most of the people. Yes, it's true.

Math, challenging it may seem and a bit mind provoking at times, is the best when it's executed with an active mind and sharp ideas. What I am referring is that illegal math becomes really hard to find sometimes.

Often people not only don't realize what happened but they are also not able to back up for how they missed that silly opening. Well you can even bet money over many of these perfectly great examples of illegal math.

So here we are counting down on our list of Top 10 Best Examples Of Illegal Math To Fool People.

This list deals with many situations and thoughts that can fool you too so buckle up as we're diving straight to the list.

#10. Infinite Money?

I mean really did they both made a profit of $10?

See, we both put $20 bills in a box, (So $40 is in the box). You give me $30 thinking that you're gonna get $40, I give you the box.

I'm gonna make $10 profit because I invested only $20 but got $30 back. But you invested a total of $20+$30=$50 and you got only $40 back.

That means you lost $10 :(

#9. Think More

Bat and a ball costs $1.10, if the bat is $1 more than the ball. How much does the ball cost?

$0.10 dollars?

Your answer is wrong then.

Because if the ball costs $0.10 then the bat would cost a dollar more (i.e 1.10$). Then the total would be $1.20 not $1.10. (Use algebra)

x is the cost of ball

x + ($1.00 + x) = $1.10

$1.00 + 2x = $1.10

2x = $1.10 – $1.00

2x = $0.10

Finally, solve for x:

x = $0.05

So, the ball must cost $0.05, and the bat must cost $1.05 since $1.05 + $0.05 = $1.10.

#8. π = 4?

It’s a nice proof because it’s logical and, even though you’ve been told that pi is 3 and a bit, the proof looks good.

It goes like this.

Start with a circle diameter 1. Thus the circumference is pi x 1, or pi.

Draw a square around it, touching it. Each edge of this square is 1, so the total is 4.

Then fold in each corner so that the corners touch the circle. The length on the left is now 1 long line and 2 smaller ones. However the total is still 1. Similar with the other edges. So the total edge is still 4, and the shape is slightly closer to the circle.

Do it again with the (now 8) outer corners. You still have a shape, edges being 4 and closer to a circle.

Each time you do this, you get closer to the circle, but the edges are still adding up to 4.

Continuing this to infinity, and the edge is now the circle, but still has a total length of 4. But we said that the circumference was pi. So pi=4.

#7. Easy But Good

Let

a = b

a = b

\Rightarrow a^{2} = a b

\Rightarrow a^{2} = a b

\Rightarrow a^{2} - b^{2} = a b - b^{2}

\Rightarrow a^{2} - b^{2} = a b - b^{2}

\Rightarrow (a - b) (a + b) = (a - b) b

\Rightarrow a + b = b

\Rightarrow 2 b = b

∴ 2 = 1

If you didn’t notice, the mistake is between the third and fourth line when you divided by (a - b), which is undefined since you are dividing by zero.

This is a common problem since it is often quite easy to disregard.

#6. Integration

Using Integration by parts, with

u = \frac{1}{x}

d v = d x

d u = - \frac{d x}{x^{2}}

v = x

\int \frac{d x}{x} = x \frac{1}{x} + \int x \frac{d x}{x^{2}}

\int \frac{d x}{x} = 1 + \int \frac{d x}{x}

Subtract

\int \frac{d x}{x}

from both sides:

0 = 1

Well, can you find what's wrong? :)

#5. Integration Again

Evaluate:

\int_{0}^{2 π} \frac{d x}{2 + \cos x}

The integral is continuous and positive everywhere, so the integral should also be positive.

\int \frac{d x}{2 + \cos x} = \frac{2}{\sqrt{3}} \arctan (\frac{\tan \frac{x}{2}}{\sqrt{3}})

And so, the definite integral is:

\int_{0}^{2 π} \frac{d x}{2 + \cos x} = {[\frac{2}{\sqrt{3}} \arctan (\frac{\tan \frac{x}{2}}{\sqrt{3}})]}_{0}^{2 π}

And if you evaluate that, you get... zero.

The correct answer, by the way, is

\frac{2 π}{\sqrt{3}}

#4. How I Proved 0/0 Is Undefined

It all started when I came across this gem:

0 \div 0 = \frac{100 - 100}{100 - 100}

0 \div 0 = \frac{10^{2} - 10^{2}}{10 (10 - 10)}

0 \div 0 = \frac{(10 + 10) (10 - 10)}{10 (10 - 10)}

0 \div 0 = \frac{10 + 10}{10}

0 \div 0 = \frac{20}{10}

0 \div 0 = 2

I was so thrilled with this proof, that I showed it to all the living beings I knew.

But not very long after, I realized I could also prove that

0 \div 0 = 3

This is how:

0 \div 0 = \frac{1000 - 1000}{1000 - 1000}

0 \div 0 = \frac{10^{3} - 10^{3}}{100 (10 - 10)}

0 \div 0 = \frac{(10 - 10) (100 + 100 + 100)}{100 (10 - 10)}

0 \div 0 = \frac{300}{100}

0 \div 0 = 3

Now, I had

0 \div 0

is equal to both

2

and

3

. Soon enough I realized that you could prove

0 \div 0

is equal to just about any natural number using a similar technique!!!

And hence, because

0 \div 0

is equal to so many numbers at the same time, it is undefined!

#3. Limits

Let

f, g

f, g

be real-valued functions. Then

\begin{aligned} (f \circ g)^{'} (x) & = lim_{h \to 0} \frac{f (g (x + h)) - f (g (x))}{h} \\ = lim_{h \to 0} \frac{f (g (x + h)) - f (g (x))}{g (x + h) - g (x)} \cdot \frac{g (x + h) - g (x)}{h} \end{aligned}

Now, let

t = g (x + h) - g (x)

. Then, we have

\begin{aligned} = lim_{h \to 0} \frac{f (g (x + h)) - f (g (x))}{g (x + h) - g (x)} \cdot \frac{g (x + h) - g (x)}{h} \\ = (lim_{t \to 0} \frac{f (g (x) + t) - f (g (x))}{t}) (lim_{h \to 0} \frac{g (x + h) - g (x)}{h}) \\ = f^{'} (g (x)) \cdot g^{'} (x) \end{aligned}

There’s only a small problem with this that t might be 0 for certain values of

h

t

There is a way to fix this so that there’s no error, but it requires significantly more detail.

#2. A Proof That $3 = 0$ $3 = 0$

I like this proof because it’s a purely algebraic one that doesn’t involve dividing by zero, and it has so many interpretations.

Let

x

be a solution to the equation

x^{2} + x + 1 = 0

Moving

x + 1

to the other side gives

x^{2} = - x - 1

Dividing by

x

gives

x = - 1 - 1 / x

. (Note that this is a legal division because

x

cannot be zero by the original equation.)

Substitute

x = - 1 - 1 / x

back in for

x

the original equation:

x^{2} + (- 1 - 1 / x) + 1 = 0

Simplifying gives

x^{2} - 1 / x = 0

, or

x^{2} = 1 / x

, and multiplying by

x

gives

x^{3} = 1

x = 1

, and plugging this back into the original equation gives

(1)^{2} + (1) + 1 = 0

, or

3 = 0

.

All of the algebra leading to

x^{3} = 1

is distraction; you can get there much more quickly by simply multiplying the original equation by

x - 1

x - 1

, which is perfectly valid. Your take on this “proof” will depend on whether you choose to work with real numbers or complex numbers.

Over real numbers, note that the original equation has no solution. This can be seen, for example, by completing the square:

x^{2} + x + 1 = {(x + \frac{1}{2})}^{2} + \frac{3}{4}

x^{2} + x + 1 = {(x + \frac{1}{2})}^{2} + \frac{3}{4}

is always positive, never zero. And indeed, the only solution to

x^{3} = 1

x = 1

, so in fact, this is a perfectly valid proof by contradiction that

x^{2} + x + 1 = 0

has no solutions.

If, however, you work over complex numbers, then the original equation

x^{2} + x + 1 = 0

does have two solutions, given by the quadratic formula, and

1

is not the only solution to

x^{3} = 1

;

there are two other solutions, namely

\frac{- 1 \pm \sqrt{- 3}}{2}

, which are precisely the solutions to the original equation. So the step going from

x^{3} = 1

x = 1

is invalid.

#1. Importance Of Difference Between Real And Imaginary

This is called iffy triangle.

This seemingly works

(1^{2} + i^{2} = 0^{2} \Rightarrow 1 - 1 = 0 ◻)

but is actually wrong because the Pythagorean theorem applies to triangles in Euclidean space with lengths

\in R_{> 0}

so you can’t really put the imaginary unit up there and you can’t have a zero length hypotenuse either.

There you have it, Top 10 Best Examples Of Illegal Math To Fool People easily.

I hope you liked this compilation of great examples of illegal math that can trick people easily, and if you did make sure to try and share this with your friends and family too.

Related Reads:
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