Nullifying Negativity in Subtraction

If B > A, then:

A - B + |A - B| = 0

AND

B - A + |B - A| = |B - A|2 or |2B - 2A|

Note: The order of A & B doesn't matter inside the absolute value sandwich. Also, both variables must be positive real numbers.

Examples:

A = 5, B = 8

5 - 8 + |5 - 8| = -3 + |-3| = -3 + 3 = 0

8 - 5 + |8 - 5| = 3 + |3| = 3 + 3 = 6 & 6 = 3 × 2

A = 19, B = 23

19 - 23 + |19 - 23| = -4 + |-4| = -4 + 4 = 0

23 - 19 + |23 - 19| = 4 + |4| = 4 + 4 = 8 & 8 = 4 × 2

It's even okay for 1 of the 2 variables to be a non-integer!

A = 1/4, B = 3

1/4 - 3 + |1/4 - 3| = -(2 + 3/4) + |-(2 + 3/4)| = -2 - 3/4 + 2 + 3/4 = 0

(The additive inverses of the numbers cancel each other out & give you zero!)

3 - 1/4 + |3 - 1/4| = 2 + 3/4 + |2 + 3/4| = 2 + 3/4 + 2 + 3/4 = 5 + 1/2 & 5 + 1/2 = (2 + 3/4)2 = 4 + 3/2 = 4 + (1 + 1/2)

(You know that 4 + 1 = 5, right?)

Since negative × positive = negative, -(x + y) = -x - y.

I like to print mixed numbers with a plus between the integer & the fraction! If the mixed number is negative, then it becomes a minus! (Otherwise, you have to put parentheses!)

It's even okay for BOTH variables to be non-integers! They don't even have to be rational!

A = π, B = √(10)

π - √(10) + |π - √(10)| = π - √(10) + √(10) - π = 0

(The additive inverses of the irrational numbers cancel each other out & give you zero!)

(Also, |A - B| = B - A or -(A - B), if B > A; -(A - B) = -A + B, too, according to the distributive plus-minus law)

√(10) - π + |√(10) - π| = (√(10) - π) + (√(10) - π) = 2√(10) - 2π or -2π + 2√(10)

(x - y = -y + x)

(√(10) > π because π = 3.14159... but √(10) = 3.16227...)

The hundredths digits of the 2 irrational numbers tells you which one is bigger. 3.16 > 3.14; therefore, √(10) > π.

Wow! The final example was a mental exercise for me! You can see the irrational sum in decimal form on a calculator!

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© Derek Cumberbatch