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Assembly Language Programming

Lesson 5: Binary Math

By Robert M (adapted by Duane Alan Hahn, a.k.a. Random Terrain)

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Page Table of Contents

Original Lesson

In lesson 4, I introduced counting with bits using the binary number system. In this lesson I will cover binary arithmetic and conclude my description of negative numbers using two's complement notation.

 

 

 

 

Binary Addition

Adding binary numbers is as easy as 1, 2, 3. In fact all you ever need to know is:

 

0+0 = 0

0+1 = 1

1+0 = 1

1+1 = 2

1+1+1 = 3

 

Adding 2 binary numbers uses the same method as adding decimal numbers. The only difference is that in decimal when you add each pair of digits, if the sum is greater than 9, you carry the 10 to the next column. In binary when the sum is greater than 1, then you carry the 2 to the next column. Except the 2 is in binary notation so its %10 instead of the decimal 2.

 

The best way to see this method is probably by example. Let's add 2 numbers in decimal, and then add the same numbers in binary.

 

 

Decimal:


        34

      + 67

------------

        ||

        |+-> 4+7 = 11 ------------------+

        |          |                    |

        |          | Carry the 10       |

        |          v                    |

        +--> 3+6 + 1 = 10 -------------+|

                       |               ||

                       | Carry the 10 ||

                       v               ||

             0+0 +     1 = 1 -------> 101 = Final answer.

             ^ ^

             | |

             +-+---- Note the leading zeros needed to finish the calculation.

Binary:

 

   %0100010 (=34)

  +%1000011 (=67)

   --------

    |||||||

    ||||||+- 0+1 = %01----------------------------+

    ||||||          |                             |

    ||||||          | No carry                    |

    ||||||          v                             |

    |||||+-- 1+1 +  0 = %10 (=2) ----------------+|

    |||||                |                       ||

    |||||           +----+  Carry the 2 (=%10)   ||

    |||||           v                            ||

    ||||+--- 0+0 +  1 = %01 --------------------+||

    ||||                 |                      |||

    ||||            +----+  No Carry            |||

    ||||            v                           |||

    |||+---- 0+0 +  0 = %00 -------------------+|||

    |||                  |                     ||||

    |||             +----+  No Carry           ||||

    |||             v                          ||||

    ||+----- 0+0 +  0 = %00 ------------------+||||

    ||                   |                    |||||

    ||              +----+  No Carry          |||||

    ||              v                         |||||

    |+------ 1+0 +  0 = %01 -----------------+|||||

    |                    |                   ||||||

    |               +----+  No Carry         ||||||

    |               v                        ||||||

    +------- 0+1 +  = = %01 ----------------+||||||

                         |                  |||||||

                         |                  vvvvvvv 

                         +- No Carry       %1100101 = 101 decimal = Final answer

Ta-da!

 

Unfortunately, I chose a poor example in that there is no case of 1 + 1 + 1 in the above example, so let's try another one.

 

Binary Addition Example 2:


   %1101111 (=111 decimal, are you confused yet :P )

  +%1111011 (=123 decimal)

   --------

    |||||||

    ||||||+- 1+1 = %10 (=2) ----------------------+

    ||||||          |                             |

    ||||||          | Carry the 2 (=%10)          |

    ||||||          v                             |

    |||||+-- 1+1 +  1 = %11 (=3) ----------------+|

    |||||                |                       ||

    |||||           +----+  Carry the 2 (=%10)   ||

    |||||           v                            ||

    ||||+--- 1+0 +  1 = %10 --------------------+||

    ||||                 |                      |||

    ||||            +----+  Carry the 2 (=%10)  |||

    ||||            v                           |||

    |||+---- 1+1 +  1 = %11 -------------------+|||

    |||                  |                     ||||

    |||             +----+  Carry the 2 (=%10) ||||

    |||             v                          ||||

    ||+----- 0+1 +  1 = %10 ------------------+||||

    ||                   |                    |||||

    ||              +----+  Carry the 2       |||||

    ||              v                         |||||

    |+------ 1+1 +  1 = %11 -----------------+|||||

    |                    |                   ||||||

    |               +----+  Carry the 2      ||||||

    |               v                        ||||||

    +------- 1+1 +  1 = %11 ----------------+||||||

                         |                  |||||||

                         +-- Carry the 2 --+|||||||

                                           ||||||||

                                           vvvvvvvv 

                                          %11101010 = 234 decimal = Final answer

                                           |

                                           v

                                           Please note that the final carry forms a

                                           new digit replacing what was a leading 

                                           zero in the numbers being added.

So much for addition, now let's look at subtraction.

 

 

 

 

 

Binary Subtraction

You may be beginning to suspect, binary subtraction is very similar to decimal subtraction in the same way that binary addition is similar to decimal addition. If so, then you are correct. Binary subtraction follows the same basic rules as decimal subtraction. The difference is that when you don't have enough to perform the subtraction of 2 digits as in decimal, then you do not borrow 10. Rather you will borrow 2.

 

So:


   0 - 0 = 0

   1 - 0 = 1

   0 - 1 = 1 borrow 2

   1 - 1 = 0

   1 - 0 - 1 = 0

   1 - 1 - 1 = 0 borrow 2

           |

           v

           This 3rd digit in the subtraction represents the borrow

           from the previous step of the subtraction.

 

As I did for addition, let's try an example.

 

Decimal Example:


     124

   - 115

     ---

     |||

     ||+-->  3 - 5 = 9 with borrow of 10 -------+

     ||                               |         | 

     ||              +---- borrow ----+         |

     ||              v                          |

     |+--->  2 - 1 - 1 = 0 --------------------+|

     |                   |                     ||

     |               +---+ no borrow           ||

     |               |                         ||

     +---->  1 - 1 - 0 = 0 -------------------+||

                         |                    |||

                         +- no borrow         vvv

                                              009 = 9 final answer.

Same Subtraction Example in Binary:


     %1111100 (= 124 decimal)

   - %1110011 (= 115 decimal)

     --------

      |||||||

      |||||||

      ||||||+---> 0 - 1     = 1 with borrow of 2 ---------+

      ||||||                                   |          |

      ||||||              +------ borrow ------+          |

      ||||||              v                               |

      |||||+----> 0 - 1 - 1 = 0 with borrow of 2 --------+|

      |||||                                    |         ||

      |||||               +------ borrow ------+         ||

      |||||               v                              ||

      ||||+-----> 1 - 0 - 1 = 0 ------------------------+||

      ||||                    |                         |||

      ||||                +---+  no borrow              |||

      ||||                v                             |||

      |||+------> 1 - 0 - 0 = 1 -----------------------+|||

      |||                     |                        ||||

      |||                 +---+  no borrow             ||||

      |||                 v                            ||||

      ||+-------> 1 - 1 - 0 = 0 ----------------------+||||

      ||                      |                       |||||

      ||                  +---+  no borrow            |||||

      ||                  v                           |||||

      |+--------> 1 - 1 - 0 = 0 ---------------------+|||||

      |                       |                      ||||||

      |                   +---+  no borrow           ||||||

      |                   v                          ||||||

      +---------> 1 - 1 - 0 = 0 --------------------+||||||

                              |                     |||||||

                              v                     vvvvvvv

                              no borrow            %0001001 = 9 decimal = final answer

Ta-da!

 

So you see that binary subtraction closely parallels decimal subtraction.

 

There is a problem, however, that we have not yet addressed. If we subtract a larger number from a smaller one, the result will be negative. In lesson 4, I began discussion of the 2's complement format for binary negative numbers. In that discussion I said that a prime advantage of 2's complement was that addition and subtraction of numbers in 2's complement resulted in the correct answer in 2's complement format. Now, let's perform a subtraction with a negative result and see what happens. No new rules are required to perform the subtraction, just follow the same procedure as if the expected result will be positive.

 

Binary Subtraction Example 2:


     %1110011 (= 115 decimal)

   - %1111100 (= 124 decimal)

     --------

      |||||||

      |||||||

      ||||||+---> 1 - 0     = 1 --------------------------+

      ||||||                  |                           |

      ||||||              +---+ no borrow                 |

      ||||||              v                               |

      |||||+----> 1 - 0 - 0 = 1 -------------------------+|

      |||||                   |                          ||

      |||||               +---+ no borrow                ||

      |||||               v                              ||

      ||||+-----> 0 - 1 - 0 = 1 with borrow of 2 -------+||

      ||||                                     |        |||

      ||||                +------ borrow ------+        |||

      ||||                v                             |||

      |||+------> 0 - 1 - 1 = 0 with borrow of 2 ------+|||

      |||                                      |       ||||

      |||                 +------ borrow ------+       ||||

      |||                 v                            ||||

      ||+-------> 1 - 1 - 1 = 1 with borrow of 2 -----+||||

      ||                                       |      |||||

      ||                  +------ borrow ------+      |||||

      ||                  v                           |||||

      |+--------> 1 - 1 - 1 = 1 with borrow of 2 ----+|||||

      |                                        |     ||||||

      |                   +------ borrow ------+     ||||||

      |                   v                          ||||||

      +---------> 1 - 1 - 1 = 1 with borrow of 2 ---+||||||

                                               |    |||||||

   Leading Zeros!         +------ borrow ------+    |||||||

     |                    v                         |||||||                         

     +----------> 0 - 0 - 1 = 1 with borrow of 2 --+|||||||

                              |                    ||||||||

                              v                    vvvvvvvv

                              BORROW!             %11110111 = -9 decimal = final answer

 

 

 

 

 

Two's Complement Revisited

The answer above is indeed the expected answer of -9 represented in 2's complement binary notation. Don't worry if you are confused, it will all become clear shortly. The most important thing to note from the above example, is that at the end of our calculation there is still a borrow. Since all that remains to subtract are leading zeros, the final borrow can never be resolved. It will produce an infinite set of leading 1's to our final answer if we keep calculating forever. This is the first clue to understanding 2's complement. Negative numbers in two's complement have an infinite number of leading 1's as opposed to positive numbers which have infinite leading zeros.

 

From this we can deduce that for all 2's complement numbers the MSB indicates the sign of the number, exactly the same as sign magnitude notation (lesson 4). If the MSB is set, then the number is negative, if the MSB is clear, then the number is positive. The difference is that 2's complement has no representation for negative 0. %10000000 in 2's complement is not negative zero. Instead it is the largest negative number that the bits other than the sign bit can represent. So %10000000 has 7 zeros that 7 bits, 2^7 = 128, so %10000000 is -128 in 2's complement.

 

 

 

 

Finding the Two's Complement

Given any positive binary number you can find its two's complement by inverting all the bits (change all 0's to 1, 1's to 0's) and then add 1.

 

For example:


   55 = %01010101

         ||||||||  Invert all bits

         vvvvvvvv

        %10101010

       +        1  Add one 

        ---------

        %10101011 = -55

 

 

Likewise we can perform the same process in reverse to convert any negative number in two's complement format to its positive equivalent.


   -55 = %10101011

        -        1  Subtract 1

         ---------

         %10101010

          ||||||||  Invert all bits.

          vvvvvvvv

         %01010101 = 55  Ta-da!

 

 

 

 

The Dreaded Overflow and Underflow!

In lesson 4, I mentioned underflow and overflow. Now I will attempt to explain what they are in detail. In lesson 1, we learned that all information in a computer is stored using bits. In lessons 2 and 3 we explored using enumerations and codes made of bits to store information like kinds of fruit. In lesson 4, we learned the standard method for counting in binary. A key point in all these lessons has been finding the number of bits needed to store the information in question. Up until now, in this lesson, I have assumed I had infinite bits to work with (leading zero's and one's were not shown but implied), but in a real computer the number of bits available is limited.

 

Therefore, the storage of each number used by your program must be restricted to a finite number of bits. Since the 650X family of processors are 8-bit processors (Each instruction works with 8 bit chunks of data at a time), we will assume you will be using multiples of 8 bits to store all your numbers. You don't have to use multiples of 8, it simply makes programming easier.

 

Earlier, I showed that the MSB of a 2's complement number is the indicator of the sign of the number. For an 8 bit number (a byte) this means that the magnitude of the number is restricted to 7 bits. There are 128 possible combinations for 7 bits. So for positive numbers we can represent values 0 through 127 with a single byte in two's complement. There is no negative zero, so for negative numbers 8 bits can represent the values -1 through -128. Therefore, 1 bytes can hold signed integers in the range from -128 to 127. Recall from lesson 4, a byte can hold UNSIGNED integers from 0 to 255.

 

Overflow and Underflow are terms meant to describe when an addition or subtraction results in an answer outside the range of values that can be represented by the number of bits being used. Overflow is when the result exceeds the maximum that the number of bits and the chosen format can represent. Underflow is when the result is less that the minimum value that the number of bits and format can represent. Specifically for 8 bits:

 

For signed 8-bit numbers:


        IF A + B is > 127 then OVERFLOW has occurred.

        IF A + B is < -128 then UNDERFLOW has occurred.

        IF A - B is > 127 then OVERFLOW has occurred.

        IF A - B is < -128 then UNDERFLOW has occurred.

For unsigned 8-bit numbers:


        IF A + B is > 255 then OVERFLOW has occurred.

        IF A - B is < 0 then UNDERFLOW has occurred.

 

 

The 650X processors include hardware support for detecting overflow and underflow during math operations. In your programs if overflow or underflow is possible, then you will need to check if it has happened and respond to correct the situation when it does. The details of code to handle overflow and underflow will have to wait for many lessons. For now just understand what it means.

 

Bonus Quiz: What are possible correct responses for a program to handle overflow or underflow?

 

 

 

 

 

Binary Multiplication and Division

Let's take some time to understand how multiplication and division are accomplished with binary numbers. All modern processors have built in support for multiplying and dividing integers, but the old 650X processors do not have specific opcodes (instructions) to multiply or divide 2 integers. That means you have to write code to perform a multiply or divide using the instructions that the processor does have. Luckily, the 650X processor family has build in support for adding and subtracting binary numbers. We will therefore use addition to implement multiplication, and subtraction to perform division.

 

 

 

 

Multiplying

Think for a moment about decimal multiplication. What does it mean to say 3 times 5?

 

One way to visualize multiplication is as an area of a rectangle, Area = width times height.

 

Example: Area = Width x height = 3 x 5 = ?


         5

       VVVVV   



     > OOOOO

   3 > OOOOO  

     > OOOOO

       |

       v

       The rectangle of O characters above represents the  
       area of a 3 by 5 rectangle. To calculate the area we
       simply count the O's. There are 15 O's, which
       is the same as 3 times 5 = 15!

Likewise, to multiply on a 650X processor, you will solve the problem by adding up the O's of an imaginary rectangle. Your program will Add 5 three times 5+5+5 = 15, or it could add 3 five times 3+3+3+3+3=15. Both ways you accomplish multiplication via repeated addition.

 

 

 

 

Division

In a similar iterative fashion, division is accomplished via repeated subtraction.

 

Example: 15 divided by 5 = ?


   15 / 5 = 15 - 5 = 10 - 5 = 5 - 5 = 0  -> Remainder = 0
                                            in this case.
                 |        |       |

                 v        v       v

                 1    +   1    +  1 = 3  ->    15 / 5 = 3

Okay, that's enough for today's lesson. Please try the following exercises. Stay tuned for a big change of pace in lesson 6 - State Machines!

 

 

 

 

 

Exercises

1.Perform the following binary additions of 2's complement numbers. Express the final result in 8 bits and indicate whether UNDERFLOW or OVERFLOW occurred.

  1. %10010100 + %01101000 = ?
  2. %00110100 + %01101111 = ?
  3. %10011100 + %11111000 = ?
  4. %01010011 + %11011101 = ?

 

2.Perform the following binary subtractions of 2's complement numbers. Express the final result in 8 bits and indicate whether UNDERFLOW or OVERFLOW occurred.

  1. %10110100 - %01001000 = ?
  2. %00110100 - %01101011 = ?
  3. %10111100 - %11011010 = ?
  4. %00010111 - %11010111 = ?

 

3.Convert the following 8-bit 2's complement numbers to their negative equivalent.

  1. %00010010
  2. %11001101
  3. %11111111
  4. %10000000

 

4.Convert the following 8-bit 2's complement numbers to their positive equivalent.

  1. %10110110
  2. %11001101
  3. %11111111
  4. %10000000

 

5.Provide a description of what would constitute OVERFLOW and UNDERFLOW for addition and subtraction involving 16-bit (word) 2's-complement numbers.

 

 

 

 

 

Answers

1.Perform the following binary additions of 2's complement numbers. Express the final result in 8 bits and indicate whether UNDERFLOW or OVERFLOW occurred.

a.%10010100 + %01101000 = ?


    %10010100 

  + %01101000

    ---------

    %11111100 -> No overflow or underflow.

 

b.%00110100 + %01101111 = ?


     11111    -> carry the 2.

    %00110100 

  + %01101111 

    ---------

    %10100011 -> Overflow occurred!
                 Adding 2 positive numbers
                 resulted in a negative.

 

c.%10011100 + %11111000 = ?


     11111     -> carry the 2.

    %10011100

  + %11111000

    --------- 

    %10010100 -> No Underflow.

 

d.%01010011 + %11011101 = ?


     11 11111  -> Carry the 2.

    %01010011 

  + %11011101

    ---------

    %00110001 -> No overflow or underflow. 

 

 

2.Perform the following binary subtractions of 2's complement numbers. Express the final result in 8 bits and indicate whether UNDERFLOW or OVERFLOW occurred.

a.%10110100 - %01001000 = ?


    %10110100 

  - %01001000

  -  1  1     -> borrow 2

    ---------

    %01001100 -> Underflow occurred!
                 (-) minus (+) should not equal a (+)

 

b.%00110100 - %01101011 = ?


    %00110100 

  - %01101011

  - 11  1 11  -> borrow 2

    ---------

    %11001001 -> No overflow or underflow.

 

c.%10111100 - %11011010 = ?


    %10111100 

  - %11011010

  - 11    1   -> borrow 2

    ---------

    %11100010 -> No overflow or underflow.

 

d.%00010111 - %11010111 = ?


    %00010111 

  - %11010111 

  - 11        -> borrow 2

    ---------

    %01000000 -> No overflow or underflow.

 

 

3.Convert the following 8-bit 2's complement numbers to their negative equivalent.

a.%00010010


    %00010010 

     ||||||||

     vvvvvvvv 

    %11101101 -> invert all bits.

  +         1 -> add 1.

    ---------

    %11101110

 

b.%11001101

   %01000101 = %10111011

 

c.%11111111

   %01111111 = %10000001

 

d.%10000000


    %00000000

     ||||||||

     vvvvvvvv

    %11111111

  +         1

    ---------

    %00000000  -> Pretty cool hey!
                  There is no negative zero
                  in two's complement! 

 

 

4.Convert the following 8-bit 2's complement numbers to their positive equivalent.

a.%10110110


    %10110110

     ||||||||

     vvvvvvvv

    %01001001

  +         1 

    ---------

    %01001010

 

b.%11001101

   %11001101 = %00110011 

 

c.%11111111

   %11111111 = %00000001

 

d.%10000000


   %10000000 = %10000000 -> Note this is the one case
   
                            that doesn't work because

                            there is no way to show

                            positive 128 in 8 bits using

                            two's complement format.

 

 

5.Provide a description of what would constitute OVERFLOW and UNDERFLOW for addition and subtraction involving 16-bit (word) 2's-complement numbers.

The number of bits used to represent a two's complement number determines the range of values that can be represented. For 8-bit numbers the range is from +127 to -128. For 16-bit numbers the range is from 32767 to -32768. Therefore overflow occurs in math with 16 bit numbers if the result of the operation is larger than 32767. Underflow occurs if the result is less than -32768.

 

 

 

Other Assembly Language Tutorials

Be sure to check out the other assembly language tutorials and the general programming pages on this web site.

 

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Lesson Links

Lesson 1: Bits!

Lesson 2: Enumeration

Lesson 3: Codes

Lesson 4: Binary Counting

Lesson 5: Binary Math

Lesson 6: Binary Logic

Lesson 7: State Machines

 

 

 

 

Useful Links

Easy 6502 by Nick Morgan

How to get started writing 6502 assembly language. Includes a JavaScript 6502 assembler and simulator.

 

 

Atari Roots by Mark Andrews (Online Book)

This book was written in English, not computerese. It's written for Atari users, not for professional programmers (though they might find it useful).

 

 

Machine Language For Beginners by Richard Mansfield (Online Book)

This book only assumes a working knowledge of BASIC. It was designed to speak directly to the amateur programmer, the part-time computerist. It should help you make the transition from BASIC to machine language with relative ease.

The Six Instruction Groups

The 6502 Instruction Set broken down into 6 groups.

6502 Instruction Set

Nice, simple instruction set in little boxes (not made out of ticky-tacky).

 

 

The Second Book Of Machine Language by Richard Mansfield (Online Book)

This book shows how to put together a large machine language program. All of the fundamentals were covered in Machine Language for Beginners. What remains is to put the rules to use by constructing a working program, to take the theory into the field and show how machine language is done.

6502 Instruction Set

An easy-to-read page from The Second Book Of Machine Language.

 

 

6502 Instruction Set with Examples

A useful page from Assembly Language Programming for the Atari Computers.

 

 

6502.org

Continually strives to remain the largest and most complete source for 6502-related information in the world.

NMOS 6502 Opcodes

By John Pickens. Updated by Bruce Clark.

 

 

Guide to 6502 Assembly Language Programming by Andrew Jacobs

Below are direct links to the most important pages.

Registers

Goes over each of the internal registers and their use.

Instruction Set

Gives a summary of whole instruction set.

Addressing Modes

Describes each of the 6502 memory addressing modes.

Instruction Reference

Describes the complete instruction set in detail.

 

 

Stella Programmer's Guide

HTMLified version.

 

 

Nick Bensema's Guide to Cycle Counting on the Atari 2600

Cycle counting is an important aspect of Atari 2600 programming. It makes possible the positioning of sprites, the drawing of six-digit scores, non-mirrored playfield graphics and many other cool TIA tricks that keep every game from looking like Combat.

 

 

How to Draw A Playfield by Nick Bensema

Atari 2600 programming is different from any other kind of programming in many ways. Just one of these ways is the flow of the program.

 

 

Cart Sizes and Bankswitching Methods by Kevin Horton

The "bankswitching bible." Also check out the Atari 2600 Fun Facts and Information Guide and this post about bankswitching by SeaGtGruff at AtariAge.

 

 

Atari 2600 Specifications

Atari 2600 programming specs (HTML version).

 

 

Atari 2600 Programming Page (AtariAge)

Links to useful information, tools, source code, and documentation.

 

 

MiniDig

Atari 2600 programming site based on Garon's "The Dig," which is now dead.

 

 

TIA Color Charts and Tools

Includes interactive color charts, an NTSC/PAL color conversion tool, and Atari 2600 color compatibility tools that can help you quickly find colors that go great together.

 

 

The Atari 2600 Music and Sound Page

Adapted information and charts related to Atari 2600 music and sound.

 

 

Game Standards and Procedures

A guide and a check list for finished carts.

 

 

Stella

A multi-platform Atari 2600 VCS emulator. It has a built-in debugger to help you with your works in progress or you can use it to study classic games. Stella finally got Atari 2600 quality sound in December of 2018. Until version 6.0, the game sounds in Stella were mostly OK, but not great. Now it's almost impossible to tell the difference between the sound effects in Stella and a real Atari 2600.

 

 

JAVATARI

A very good emulator that can also be embedded on your own web site so people can play the games you make online. It's much better than JStella.

 

 

batari Basic Commands

If assembly language seems a little too hard, don't worry. You can always try to make Atari 2600 games the faster, easier way with batari Basic.

 

 

Back to Top

 

 

In Case You Didn't Know

 

Trump's Jab = Bad

Did you know that Trump's rushed experimental rona jab has less than one percent overall benefit? It also has many possible horrible side effects. Some brainwashed rona jab cultists claim that there are no victims of the jab, but person after person will post what the jab did to them, a friend, or a family member on web sites such as Facebook and Twitter and they'll be lucky if they don't get banned soon after. Posting the truth is “misinformation” don't you know. Awakened sheep might turn into lions, so powerful people will do just about anything to keep the sheep from waking up.

 

Check out these videos:

What is causing the mysterious self-assembling non-organic clots?

If You Got the COVID Shot and Aren't Injured, This May Be Why

Full Video of Tennessee House of Representatives Health Subcommittee Hearing Room 2 (The Doctors Start Talking at 33:28)

 

 

H Word and I Word = Good

Take a look at my page called The H Word and Beyond. You might also want to look at my page called Zinc and Quercetin. My sister and I have been taking those two supplements since summer of 2020 in the hopes that they would scare away the flu and other viruses (or at least make them less severe).

 

 

B Vitamins = Good

Some people appear to have a mental illness because they have a vitamin B deficiency. For example, the wife of a guy I used to chat with online had severe mood swings which seemed to be caused by food allergies or intolerances. She would became irrational, obnoxious, throw tantrums, and generally act like she had a mental illness. The horrid behavior stopped after she started taking a vitamin B complex. I've been taking Jarrow B-Right (#ad) for many years. It makes me much easier to live with.

 

 

Soy = Bad

Unfermented soy is bad! “When she stopped eating soy, the mental problems went away.” Fermented soy doesn't bother me, but the various versions of unfermented soy (soy flour, soybean oil, and so on) that are used in all kinds of products these days causes a negative mental health reaction in me that a vitamin B complex can't tame. The sinister encroachment of soy has made the careful reading of ingredients a necessity.

 

 

Wheat = Bad

If you are overweight, have type II diabetes, or are worried about the condition of your heart, check out the videos by Ken D Berry, William Davis, and Ivor Cummins. It seems that most people should avoid wheat, not just those who have a wheat allergy or celiac disease. Check out these books: Undoctored (#ad), Wheat Belly (#ad), and Eat Rich, Live Long (#ad).

 

 

Negative Ions = Good

Negative ions are good for us. You might want to avoid positive ion generators and ozone generators. A plain old air cleaner is better than nothing, but one that produces negative ions makes the air in a room fresher and easier for me to breathe. It also helps to brighten my mood.

 

 

Litterbugs = Bad

Never litter. Toss it in the trash or take it home. Do not throw it on the ground. Also remember that good people clean up after themselves at home, out in public, at a campsite and so on. Leave it better than you found it.

 

 

Climate Change Cash Grab = Bad

Seems like more people than ever finally care about water, land, and air pollution, but the climate change cash grab scam is designed to put more of your money into the bank accounts of greedy politicians. Those power-hungry schemers try to trick us with bad data and lies about overpopulation while pretending to be caring do-gooders. Trying to eliminate pollution is a good thing, but the carbon footprint of the average law-abiding human right now is actually making the planet greener instead of killing it.

 

Watch these two YouTube videos for more information:

CO2 is Greening The Earth

The Climate Agenda

 

 

How to Wake Up Normies

Charlie Robinson had some good advice about waking up normies (see the link to the video below). He said instead of verbally unloading or being nasty or acting like a bully, ask the person a question. Being nice and asking a question will help the person actually think about the subject.

 

Interesting videos:

Charlie Robinson Talks About the Best Way to Wake Up Normies

Georgia Guidestones Explained

The Men Who Own Everything

Disclaimer

View this page and any external web sites at your own risk. I am not responsible for any possible spiritual, emotional, physical, financial or any other damage to you, your friends, family, ancestors, or descendants in the past, present, or future, living or dead, in this dimension or any other.

 

Use any example programs at your own risk. I am not responsible if they blow up your computer or melt your Atari 2600. Use assembly language at your own risk. I am not responsible if assembly language makes you cry or gives you brain damage.

 

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