by Lawrence J. J. Leonard

This is how to prove 4 = 3. If you are a Math Geek, kinda like me, then you will enjoy this trick to fool your friends. *There is always a catch, though.*

Let’s assume this expression is true:** a + b = c**

This can also be re-written as:

**4a – 3a + 4b – 3b = 4c – 3c**

So far, so good. All true.

Now **Reorganize**:

**[-4c]** 4a – 3a + 4b – 3b = 4c – 3c **[-4c]**

4a – 3a + 4b – 3b – 4c = -3c .Okay.

**[+3a+3b**] 4a – 3a + 4b – 3b – 4c = -3c **[+3a+3b]**

4a + 4b – 4c = 3a + 3b -3c .Okay.

**Remove common factors**:

**[4] ** 4a + 4b – 4c = 3a + 3b – 3c **[3]**

**This is the result:**

4 (a+b-c) = 3 (a+b-c)

Extract the Common Terms on both sides.

They cancel each other out.

**Therefore:**

** 4 = 3**

**NOT SO FAST !**

This is known as a “Howler” or blunder because it is based on a *fallacy of multiplication* — The coefficients, in this case 4 and 3, are dependent on what is in the brackets.

If a + b = c, then when you ‘remove common factors,’

what is in the brackets will equal zero — because of the “-c” (negative c).

Remember the equation you are left with is, in actuality, this: 4(0) = 3(0)

This is true only if you multiply the coefficients.

This is true without the coefficients.

This is **not true** if you ignore the **PEMDAS** rule, and get 4=3. (Ooooo!)

How Do I remember the **PEMDAS **rule?

**P** – Parentheses first

**E** – Exponents (i.e. Powers and Square Roots, etc.)

**MD **– Multiplication and Division (left-to-right)

**AS** – Addition and Subtraction (left-to-right)

Copyright © 1960-2015 Lawrence J. J. Leonard All rights reserved

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