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As an engineer, I have had numerous high school and college math classes:

- Algebra I
- Algebra II
- Geometry
- Trigonometry
- Calculus I
- Calculus II
- Calculus III
- Differential Equations
- Numerical Methods

Yet, I never cease to be amazed by the so-called "new math". One of the things that threw me for a loop the other day is when my kid asked me for an explanation for this factoring of a binomial. He had written down every step the teacher had written on the board, but was completely befuddled by the what he had copied verbatim. And quite honestly, I did not understand it either because it seemingly violates the laws of algebra, yet allegedly produces the correct answer:

**Line 1:** $$ 3x^2 - 17x + 20 $$

**Line 2:** $$ x^2 - 17x + 60 $$

**Line 3:** $$ (x - 5)(x - 12) $$

**Line 4:** $$ \left( x - \frac{5}{3} \right) \left( x - \frac{12}{3} \right) $$

**Line 5:** $$ (3x - 5)(x - 4) $$

Some of that makes sense. But, the reason I say it "seemingly violates the laws of algebra" is because **there is not a way using algebra to get from Line 1 to Line 2. Similarly, there is no way to get from Line 3 to Line 4 using algebra**. Based on the *distributive property* ...

- If Line 1 is divided by 3 to remove the coefficient 3 from the first addend 3x
^{2}then the second and third addends -17x and 20 must also be divided by 3, but they are not. - If Line 1 is multiplied by 3 to triple the third addend 20 then the first and second addends 3x
^{2}and -17x must also be multiplied by 3, but they are not. - If Line 3 is divided by 3
^{2}such that (x - 5) and (x - 12) are each divided by 3, then the x variables should also be divided by 3, but they are not.

As it turns out, the steps in the example above are not actually algebraic steps. Rather, **they are steps of an algorithm**. If all of the steps of the algorithm are executed correctly, then the algorithm yields the correct result. This algorithm goes by several different names, with the more common names shown below:

- Slide and Divide
- Swing and Divide
- Slip and Slide

After some deep thought on this apparent conundrum, I concluded that **the variable x that exists on Lines 1, 4 & 5 is not actually the same variable x that exists on Line 2 & 3**. That is why there is not an algebraic way to move from Line 1 to Line 2, or from Line 3 to Line 4! But, the steps of the algorithm can be reconciled with algebra by changing the x variable on Lines 2 & 3 to a new variable, named T for "temporary":

**Line 1:** Original Binomial

$$ 3x^2 - 17x + 20 $$

**Line 2:** Slide the 3 and Change the Variable Name

$$ T^2 - 17T + 60 $$

**Line 3:** Factor the Binomial

$$ (T - 5)(T - 12) $$

**Line 4:** Divide by 3 and Restore the Variable Name

$$ \left( x - \frac{5}{3} \right) \left( x - \frac{12}{3} \right) $$

**Line 5:** Simplify to Standard Form

$$ (3x - 5)(x - 4) $$

Once the temporary variable T is used, the lines above are actually algebra with some of the intermediate steps removed. The rest of this post is dedicated to "show all the work" so people can understand why it works instead of just blindly trusting in a confusing magic sequence of steps.

In order to factor the following binomial (which is assumed to equal zero)

Line 1:$$ 3x^2 - 17x + 20 $$

**First**, divide by 3 to remove the coefficient of the first addend, and simplify

$$ \frac{3x^2 - 17x + 20}{3} $$

$$ \frac{3x^2}{3} - \frac{17x}{3} + \frac{20}{3} $$

$$ x^2 - \frac{17}{3}x + \frac{20}{3} $$

**Second**, define variable x in terms of a new, temporary variable T divided by the original coefficient of the first addend.

$$ x = \frac{T}{3} $$

**Third**, substitute the new definition of x back into the equation, and simplify

$$ x^2 - \frac{17}{3}x + \frac{20}{3} $$

$$ \left ( \frac{T}{3} \right )^2 - \left( \frac{17}{3} \right) \left(\frac{T}{3} \right) + \frac{20}{3} $$

$$ \frac{T^2}{3^2} - \frac{17T}{3^2} + \frac{20}{3} $$

**Fourth**, multiple by 3^{2} to remove the fractions on the first and second addends in the expression above, and simplify.

$$ 3^2 \left( \frac{T^2}{3^2} - \frac{17T}{3^2} + \frac{20}{3} \right) $$

$$ 3^2 \left( \frac{T^2}{3^2} \right) - 3^2 \left( \frac{17T}{3^2} \right) + 3^2 \left( \frac{20}{3} \right) $$

$$ T^2 - 17T + 3 \bullet 20 $$

Line 2:$$ T^2 - 17T + 60 $$

There it is: **the slide**:

- The original coefficient, 3, of the first addend has been removed from the first addend
- Nothing has been done to the original coefficient, -17, of the second addend
- The original coefficient, 3, of the first addend has been multiplied by the original coefficient, 20, of the third addend

**Fifth**, factor the new binomial that is written in terms of T.

$$ T^2 - 17T + 60 $$

Line 3:$$ (T - 5)(T - 12) $$

**Sixth**, find the definition of T, and substitute it back into the equation above:

$$ x = \frac{T}{3} $$

$$ T = 3x $$

$$ (T - 5)(T - 12) $$

$$ (3x - 5)(3x - 12) $$

**Seventh**, divide the expression above by 3^{2} to remove the coefficient, 3, from both terms, and simplify.

$$ (3x - 5)(3x - 12) $$

$$ \frac{(3x - 5)(3x - 12)}{3^2} $$

$$ \left( \frac{3x - 5}{3} \right) \left( \frac{ 3x - 12}{3} \right) $$

Line 4:$$ \left( x - \frac{5}{3} \right) \left( x - \frac{12}{3} \right) $$

There it is: **the divide**: The factors obtained after doing the slide, 5 and 12, have been divided by the original coefficient, 3, of the first addend.

**Finally**, the factorization can be simplified into common form:

]]>

Line 5:$$ (3x - 5)(x - 4) $$

Obviously, I can expand the number of USB 3.0 ports via an external hub or a cheap USB 3.0 PCIe card with more ports (I have seen as many as 7-ports on one card), but that would keep all of the USB 3.0 ports on the same channel. Its seem to have two options to create an additional USB 3.0 channel:

* I can replace the existing 4-port USB 3.0 card with a card that has two or more "dedicated" channels

* I can replace my existing Gigabit network card with a new card that as USB 3.0 in addition to Gigabit network

The first option tends to be more pricey, and reduces fault tolerance by connecting both USB drives onto the same physical card. The second option is generally a bit cheaper, and increases fault tolerance by putting both drives onto separate USB channels. If one card fails or has a driver issue, the other card my still work so I night still have access to 1/2 of the Storage Space mirror. I decided to go with the second option.

I found three options, organized from lowest price to highest price:

* Syba SD-PEX50100

* SIIG LB-US0614-S1

* StarTech PEXUSB3S3GE

All three of these cards have:

* 1x Gigabit ethernet port

* 3x USB 3.0 (aka USB 3.1 Gen 1, 5GHz Full-Duplex) ports on one channel

All three of these cards have specs that say they support Windows 10, 64-bit.

The Syba and StarTech cards use a SATA power connector, whereas the SIIG card uses an old-school 5-pin Molex power connector.

The Syba and SIIG cards have the USB ports oriented vertically on the backplane connector, whereas the StarTech card has the USB ports oriented horizontally on the backplane connector. I personally prefer the vertical orientation because that tends to make the ports a bit easier to plug and unplug cables.

I have only had one Syba produce, and did not have much luck with it's reliability. However, I have owned multiple SIIG and StarTech products and have never had an issue with any of them.

I decided on the SIIG product, and will come back and update this post with any relevant information after I have it installed.

]]>Just what is the "number base" of a number system? Simply put, it's the number of unique characters that represent a single digit position of a multi-digit number. For example, each decimal digit is a number between 0 - 9. Computers commonly use bits (abbreviation for "binary digit"), octal digits and hex digits.

Number System | Base | Digits |
---|---|---|

Decimal | 10 | 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 |

Binary | 2 | 0, 1 |

Octal | 8 | 0, 1, 2, 3, 4, 5, 6, 7 |

Hexadecimal (Hex) | 16 | 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F |

All readers are familiar with the decimal number system. You learned it as a small child when you learned how to count. There is a "one's digit", "ten's digit", "hundred's digit", "thousand's digit",and so on. The least significant digit goes on the right, and the digits increase in significance (i.e. value) as you move to the left. In math itself, there is no limit to the number of digit positions that can be included to the left.

You probably never stopped to think about why the digits are called the "one's digit", "ten's digit", "hundred's digit","thousand's digit", etc.. Mathematically speaking, it's really pretty straight forward: because there are ten unique characters per digit position, they represent ten (i.e base-10) raised to the power of the indexed position, starting with 0 on the right ( ... 3, 2, 1, 0).

Tousand's Digit | Hundred's Digit | Ten's Digit | One's Digit |
---|---|---|---|

10^{3} = 1000 |
10^{2} = 100 |
10^{1} = 10 |
10^{0} = 1 |

Counting follows a well-known pattern:

- There is fixed number of characters per digit (i.e. 0 - 9).
- Before you start counting, all digits start at the first character: 0
- Every time you increment, you advance the least significant digit (one's digit) by 1 character until all of the characters are used for that digit: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
- Once you have used all of the characters at that digit, you increment the next more significant digit (i.e. to the left) by 1 character, and reset the current digit's character to 0.

Hundred's Digit | Ten's Digit | One's Digit | Comments |
---|---|---|---|

0 | 0 | 0 | All digits start at zero |

0 | 0 | 1 | |

0 | 0 | 2 | |

0 | 0 | 3 | |

0 | 0 | 4 | |

0 | 0 | 5 | |

0 | 0 | 6 | |

0 | 0 | 7 | |

0 | 0 | 8 | |

0 | 0 | 9 | Last character for the one's digit position |

0 | 1 | 0 | Reset least significant digit to zero; Increment next digit to left by one |

0 | 1 | 1, 2, 3, 4, 5, 6, 7, 8 | |

0 | 1 | 9 | Last character for one's digit position |

0 | 2 | 0 | Reset least significant digit to zero; Increment next digit to left by one |

0 | ... | ... | |

0 | 9 | 9 | Last character for one's digit position; Also last digit for ten's digit position |

1 | 0 | 0 | Reset one's digit and hundred's digit to zero; Increment next digit to left by one |

Counting in binary follow the same rules as counting in decimal. The only difference is that in decimal there are ten characters per digit (i.e. 0 - 9), whereas in binary there are only two characters per digit (i.e. 0 - 1). Than means instead of a "one's digit", "ten's digit", "hundred's digit", etc. there is a "one's digit", "two's digit", "four's digit", "eight's digit", etc. In other words, instead of each more significant digit being a power of ten as in decimal, each more significant digit is instead a power of two in binary.

Eight's Digit | Four's Digit | Two's Digit | One's Digit |
---|---|---|---|

2^{3} = 8 |
2^{2} = 4 |
2^{1} = 2 |
2^{0} = 1 |

But, in math, counting is counting regardless of number base, so counting in binary follows the same rules as counting in decimal, with the only difference being the number of characters per digit:

- There is fixed number of characters per digit (i.e. 0 - 1).
- Before you start counting, all digits start at the first character: 0
- Every time you increment, you advance the least significant digit (one's digit) by 1 character until all of the characters are used for that digit: 1
- Once you have used all of the characters at that digit, you increment the next more significant digit (i.e. to the left) by 1 character, and reset the current digit's character to 0.

Four's Digit | Two's Digit | One's Digit | Comments |
---|---|---|---|

0 | 0 | 0 | All digits start at zero |

0 | 0 | 1 | Last character for one's digit position |

0 | 1 | 0 | Reset least significant digit to zero; Increment next digit to left by one |

0 | 1 | 1 | Last character for one's digit position; Also last character for two's digit position |

1 | 0 | 0 | Reset one's digit and two's digit to zero; Increment the next digit to the left by one |

Of course, with a total of ten fingers, you could actually count from 0 -1023 in binary using your fingers!

512 | 256 | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
---|---|---|---|---|---|---|---|---|---|

1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |

Counting in binary on your fingers is not very practical, though; I am pretty sure most people are not dexterous to make all 1024 (0 - 1023) of the finger combinations. However, binary is particularly well-suited for computers which uses transistor that can be switched into one of two states: "on" or "off".

By now, you should understand that the value of a number is simply the digit at a specific place position times the value of the place position. Consider, for example, the decimal number 9876. That is equal to

9 * 10^{3} + 8 * 10^{2} + 7 * 10^{1} + 6 * 10^{0} = 9 * 1000 + 8 * 100 + 7 * 10 + 6 * 1 = 9000 + 800 + 70 + 6 = 9876

10^{3} = 1000 |
10^{2} = 100 |
10^{1} = 10 |
10^{0} = 1 |
---|---|---|---|

9 | 8 | 7 | 6 |

Binary is no different. Consider the number binary number 1001. The decimal value of that binary number can be computed the same way:

1 * 2^{3} + 0 * 2^{2} + 0 * 2^{1} + 1 * 2^{0} = 1 * 8 + 0 * 4 + 0 * 2 + 1 * 1 = 8 + 0 + 0 +1 = 9

2^{3} = 8 |
2^{2} = 4 |
2^{1} = 2 |
2^{0} = 1 |
---|---|---|---|

1 | 0 | 0 | 1 |

Binary digits, or bits, are commonly grouped into packs of bits:

Number of bits | Nomenclature |
---|---|

1 | bit |

4 | Nibble |

8 | Byte |

Since a Nibble is 4 bits, that means that a nibble can represent binary values from 0000 to 1111. That is where hexadecimal comes in. Since hexadecimal is base-16, that means it has 16 different symbols (0 - 9, A -F) so if fits perfectly with a Nibble.

Binary | Hexadecimal | Decimal |
---|---|---|

0000 | 0 | 0 |

0001 | 1 | 1 |

0010 | 2 | 2 |

0011 | 3 | 3 |

0100 | 4 | 4 |

0101 | 5 | 5 |

0110 | 6 | 6 |

0111 | 7 | 7 |

1000 | 8 | 8 |

1001 | 9 | 9 |

1010 | A | 10 |

1011 | B | 11 |

1100 | C | 12 |

1101 | D | 13 |

1110 | E | 14 |

1111 | F | 15 |

Therefore, a Byte, which is an 8-bits, can also be represented by two hexadecimal digits between 00 - FF, which has a decimal value of 0 - 255.

Binary | Hexadecimal | Decimal |
---|---|---|

0000 0000 | 00 | 000 |

0000 0001 | 01 | 001 |

0000 0010 | 02 | 002 |

... | ... | ... |

0000 0100 | 04 | 004 |

... | ... | ... |

0000 1000 | 08 | 008 |

... | ... | ... |

0001 0000 | 10 | 16 |

... | ... | ... |

0010 0000 | 20 | 32 |

... | ... | ... |

0100 0000 | 40 | 64 |

... | ... | ... |

1000 0000 | 80 | 128 |

... | ... | ... |

1000 1000 | 88 | 136 |

... | ... | ... |

11111111 | FF | 255 |