commit: 1bb7876c46c0d3736006ba16d62ab5eb445e889f
parent: 3ca46d3a9ebbee316522963ff90cc7c24bee1967
author: Chris Noxz <chris@noxz.tech>
date: Mon, 15 Aug 2022 11:02:09 +0200
add seminew article about divisibility theorems
6 files changed, 97 insertions(+), 109 deletions()
diff git a/noxz.tech/guides/mathematics/.assemble b/noxz.tech/articles/divisibility_theorems_and_fraction_flipping/.assemble
rename noxz.tech/guides/mathematics/.assemble → noxz.tech/articles/divisibility_theorems_and_fraction_flipping/.assemble
diff git a/noxz.tech/articles/divisibility_theorems_and_fraction_flipping/.buildignore b/noxz.tech/articles/divisibility_theorems_and_fraction_flipping/.buildignore
diff git a/noxz.tech/articles/divisibility_theorems_and_fraction_flipping/.metadata b/noxz.tech/articles/divisibility_theorems_and_fraction_flipping/.metadata
@@ 0,0 +1,5 @@
+.ds YEAR 2019
+.ds MONTH September
+.ds DAY 21
+.ds DATE \*[MONTH] \*[DAY], \*[YEAR]
+.ds TITLE Divisibility theorems and fraction flipping
diff git a/noxz.tech/guides/mathematics/index.www b/noxz.tech/articles/divisibility_theorems_and_fraction_flipping/index.www
rename noxz.tech/guides/mathematics/index.www → noxz.tech/articles/divisibility_theorems_and_fraction_flipping/index.www
@@ 1,10 +1,4 @@
.\".fam C
.\"
Under this topic I will not just show tricks, as the notion of tricks in
mathematics is probably the most damning to the spread of mathematical
knowledge. I will instead explain how these, so called, tricks work using
logic and reason. So this topic is not for the savvy hackers or mathematicians
out there.
.DIVS toc
.SPAN toctitle Contents
@@ 63,15 +57,15 @@ out there.
Divisibility theorems
.HnE
When it comes to divisibility there exists some neat theorems to test a certain
number's different divisibilities, or factors. Following are those theorems and
their proof. Some of the proofs are more trivial than others, such as the proof
for divisibility by 1, 2, 5 and 10. From here on out it's assumed that every
number, when working with divisibility, is an integer.
+When it comes to divisibility, there exist some neat theorems to test a certain
+number’s different divisibilities, or factors. The following are those theorems
+and their proofs. Some of the proofs are more trivial than others, such as the
+proof for divisibility by 1, 2, 5, and 10. From here on out, it’s assumed that
+every number, when working with divisibility, is an integer.
What is the smallest number that is divisible by 1 through 10? The answer is
.SPAN spoiler 2520 .
You can try the different divisibility theorems below on it.
+You can try the different divisibility theorems below to prove it.
.HnS 2
.TAG "mathdivtheorems1"
@@ 92,9 +86,9 @@ Divisibility by 2
.HnE
It's common knowledge that every even number (numbers ending with an even
number) is divisible by 2. This is because even numbers are multiples of 2. In
short, if a number ends with either 0, 2, 4, 6 or 8 it is divisible by 2.
+short, if a number ends with either 0, 2, 4, 6, or 8, it is divisible by 2.
Say we have a four digit number
+Say we have a fourdigit number
.I abcd ,
where
.I a
@@ 102,9 +96,9 @@ represents the number of thousands,
.I b
the number of hundreds,
.I c
the number of tens and
.I d
the number of ones. This number can be represented as
+the number of tens, and the number
+.I d .
+This number can be represented as
.I a "" 1000
+
.I b "" 100
@@ 114,7 +108,7 @@ the number of ones. This number can be represented as
.I d .
If
.I d
is divisible by 2 we can represent it as an even number
+is divisible by 2, we can represent it as an even number
.I n , 2
like so:
@@ 148,7 +142,7 @@ the whole number is divisible by 3, i.e.
Why is this proposition true?
Let's use the four digit number
+Let's use the fourdigit number
.I abcd ,
where
.I a
@@ 156,9 +150,9 @@ represents the number of thousands,
.I b
the number of hundreds,
.I c
the number of tens and
.I d
the number of ones. this number can be represented as
+the number of tens, and the number
+.I d .
+This number can be represented as
.I a "" 1000
+
.I b "" 100
@@ 168,7 +162,7 @@ the number of ones. this number can be represented as
.I d .
Now let's break out one
.I a ,
.I b
+.I b ,
and
.I c
from the first 3 terms, and factor out 3 like so:
@@ 201,16 +195,16 @@ is divisible by 3. And so the theorem is proven.
.DIVE
We have shown that the procedure above will hold for all cases. The procedure
is also recursive. If the sum is to hard to test divisibility for, the
procedure can be repeated, until a smaller sum is reveald. This is rarely
necessary as the divisibility of the sum often is easy to determined.
+is also recursive. If the sum is too hard to test for divisibility, the
+procedure can be repeated until a smaller sum is revealed. This is rarely
+necessary as the divisibility of the sum is often easy to determine.
.HnS 2
.TAG "mathdivtheorems4"
Divisibility by 4
.HnE
The theorem goes that if the last two digits of a number is divisible by 4, the
whole number is divisible by 4, i.e.
+The theorem goes that if the last two digits of a number are divisible by 4,
+the whole number is divisible by 4, i.e.
.EQ
4 ~~ abcd ~iff~ 4 ~~ cd
@@ 218,7 +212,7 @@ whole number is divisible by 4, i.e.
Why is this proposition true?
Let's use the four digit number
+Let's use the fourdigit number
.I abcd ,
where
.I a
@@ 226,9 +220,9 @@ represents the number of thousands,
.I b
the number of hundreds,
.I c
the number of tens and
.I d
the number of ones. This number can be represented as
+the number of tens, and the number
+.I d .
+This number can be represented as
.I a "" 1000
+
.I b "" 100
@@ 260,7 +254,7 @@ theorem is proven.
.TAG "mathdivtheorems5"
Divisibility by 5
.HnE
The theorem goes that if the last digits of a number is divisible by 5, the
+The theorem goes that if the last digits of a number are divisible by 5, the
whole number is divisible by 5, i.e.
.EQ
@@ 269,7 +263,7 @@ whole number is divisible by 5, i.e.
Why is this proposition true?
Let's use the four digit number
+Let's use the fourdigit number
.I abcd ,
where
.I a
@@ 277,9 +271,9 @@ represents the number of thousands,
.I b
the number of hundreds,
.I c
the number of tens and
.I d
the number of ones. This number can be represented as
+the number of tens, and the number
+.I d .
+This number can be represented as
.I a "" 1000
+
.I b "" 100
@@ 299,9 +293,9 @@ above
We can now see that the first term is divisible by 5, so the whole number is
divisible by 5 if, and only if, the second term is divisible by 5. And so the
theorem is proven. As the only one digit numbers that are divisible by 5 are 0
and 5, another way of putting it is \(en if last digit is 0 or 5, the number is
divisible by 5.
+theorem is proven. As the only onedigit numbers that are divisible by 5 are 0
+and 5, another way of putting it is that if the last digit is 0 or 5, the
+number is divisible by 5.
.DIVS theorem
.EQ
@@ 325,14 +319,14 @@ and
.DIVE
.B Note :
Be careful when combining theroems like this. Don't be fooled and try
to combine divisibility theorems for 2 and 4 to get the theorem for 8, as
numbers divisible by 4 always are divisible by 2. The number 4 is for instance
both divisible by 2 and 4, but
+Be careful when combining theorems like this. Don't be fooled and try to
+combine the divisibility theorems for 2 and 4 to get the theorem for 8, as
+numbers divisible by 4 are always divisible by 2. The number 4 is, for
+instance, both divisible by 2 and by 4, but
.I not
by 8. Make sure the theorems you combine
doesn't share a factor, like 4 and 2 sharing the factor 2. It's of course
possible to use theoerems 4 and 2 together, but it's not certain in all cases
+doesn't share a factor, like 4 and 2 sharing factor 2. It's, of course,
+possible to use theorems 4 and 2 together, but it's not certain in all cases
that the two theorems prove divisibility by 8.
.HnS 2
@@ 343,9 +337,9 @@ Probably one of the most useful theorems is the theorem of
.I "divisibility by 7" ,
as it is recursive (just like the theorems of
.I "divisibility by 3 & 9" ).
The theorem states that if the difference between the last digit multiplied by 2
and the remaining digits in a number is divisible by 7, the whole number is
divisible by 7. I'll show the procedure with an example below:
+The theorem states that if the difference between the last digit multiplied by
+2 and the remaining digits in a number is divisible by 7, the whole number is
+divisible by 7. I'll show you the procedure with an example below:
.EQ
lpile {
@@ 359,7 +353,7 @@ above
}
.EN
Neat! So how and why does it work? For simplicity's sake we use a two digit
+Neat! So how and why does it work? For simplicity's sake, we use a twodigit
number
.I ab ,
represented as
@@ 391,13 +385,13 @@ If we have

.I b , 2
and it's divisible by 7, we know that 7 must be a factor of the
expression. We can now create an equation:
+expression. We can now create an equation.
.EQ
{ a ~~ 2b ~=~ 7k }
.EN
Multiply the whole equation with 10, and add one extra
+Multiply the whole equation by 10, and add one extra
.I b :
.EQ
@@ 429,12 +423,13 @@ left side says
.I a "" 10
+
.I b .
Neat, we now know that the two digit number
+Neat, we now know that the twodigit number
.I ab
is divisible by 7. Now we must show that
.B B implies
+.B B
+implies
.B A .
That is if
+That is, if
.I a

.I b "" 2
@@ 469,12 +464,12 @@ above
.EN
We can now see that the right side of the equation is divisible by 7, and on
our left side 10 is not divisible by 7 so the expression inside the
+our left hand, 10 is not divisible by 7, so the expression inside the
parenthesis must be. But isn't that expression
.I a

.I b . 2
Neat, we now have the proof for the theorem and can conclude that indeed:
+.I b ? 2
+We now have the proof for the theorem and can conclude that, indeed,
.DIVS theorem
.EQ
@@ 490,9 +485,8 @@ Divisibility by 8
.HnE
The theorem is quite similar to the theorem for
.URL #mathdivtheorems4 "divisibility by 4" .
The theorem goes that if the last three digits of a number is divisible by 8,
the
whole number is divisible by 8, i.e.
+The theorem goes that if the last three digits of a number are divisible by 8,
+then the whole number is divisible by 8, i.e.
.EQ
8 ~~ abcd ~iff~ 8 ~~ bcd
@@ 500,7 +494,7 @@ whole number is divisible by 8, i.e.
Why is this proposition true?
Let's use the four digit number
+Let's use the fourdigit number
.I abcd ,
where
.I a
@@ 508,9 +502,9 @@ represents the number of thousands,
.I b
the number of hundreds,
.I c
the number of tens and
.I d
the number of ones. This number can be represented as
+the number of tens, and the number
+.I d .
+This number can be represented as
.I a "" 1000
+
.I b "" 100
@@ 518,7 +512,7 @@ the number of ones. This number can be represented as
.I c "" 10
+
.I d .
Now let's factor out 8 from the first term, like so:
+Now let's factor out 8 from the first term, like so,
.EQ
lpile {
@@ 545,7 +539,7 @@ Divisibility by 9
Much like the theorem for
.URL #mathdivtheorems3 "divisibility by 3" ,
the theorem goes that if the sum of all digits in a number is divisible by 9,
the whole number is divisible by 9, i.e.
+the whole number is divisible by 9, i.e.,
.EQ
9 ~~ abcd ~iff~ 9 ~~ left ( a ~+~ b ~+~ c ~+~ d right )
@@ 553,7 +547,7 @@ the whole number is divisible by 9, i.e.
Why is this proposition true?
Let's use the four digit number
+Let's use the fourdigit number
.I abcd ,
where
.I a
@@ 561,9 +555,9 @@ represents the number of thousands,
.I b
the number of hundreds,
.I c
the number of tens and
.I d
the number of ones. This number can be represented as
+the number of tens, and the number
+.I d .
+This number can be represented as
.I a "" 1000
+
.I b "" 100
@@ 571,7 +565,7 @@ the number of ones. This number can be represented as
.I c "" 10
+
.I d .
Now let's break out one a, b and c from the first 3 terms, and factor out 9
+Now let's break out one a, b, and c from the first 3 terms, and factor out 9
like so:
.EQ
@@ 602,15 +596,15 @@ is divisible by 9. And so the theorem is proven.
.DIVE
We have shown that the procedure above will hold for all cases. The procedure
is also recursive. If the sum is to hard to test divisibility for, the
procedure can be repeated, until a smaller sum is reveald. This is rarely
necessary as the divisibility of the sum often is easy to determined.
+is also recursive. If the sum is too hard to test for divisibility, the
+procedure can be repeated until a smaller sum is revealed. This is rarely
+necessary as the divisibility of the sum is often easy to determine.
.HnS 2
.TAG "mathdivtheorems10"
Divisibility by 10
.HnE
The theorem goes that if the last digits of a number is divisible by 10, the
+The theorem goes that if the last digits of a number are divisible by 10, the
whole number is divisible by 10, i.e.
.EQ
@@ 619,7 +613,7 @@ whole number is divisible by 10, i.e.
Why is this proposition true?
Let's use the four digit number
+Let's use the fourdigit number
.I abcd ,
where
.I a
@@ 627,9 +621,9 @@ represents the number of thousands,
.I b
the number of hundreds,
.I c
the number of tens and
.I d
the number of ones. This number can be represented as
+the number of tens, and the number
+.I d .
+This number can be represented as
.I a "" 1000
+
.I b "" 100
@@ 649,8 +643,8 @@ above
We can now see that the first term is divisible by 10, so the whole number is
divisible by 10 if, and only if, the second term is divisible by 10. And so the
theorem is proven. As the only one digit number that is divisible by 10 is 0,
another way of putting it is \(en if last digit is 0, the number is
+theorem is proven. As the only onedigit number that is divisible by 10 is 0,
+another way of putting it is that if the last digit is 0, the number is
divisible by 10.
.DIVS theorem
@@ 664,7 +658,7 @@ divisible by 10.
Divisibility by 11
.HnE
The theorem goes that a number is divisible by 11 if, and only if, the
alternate sum of its digits is divisible by 11, like so:
+alternate sum of its digits is divisible by 11. Like so,
.EQ
lpile {
@@ 676,7 +670,7 @@ above
}
.EN
Neat! So how and why does it work? For simplicity's sake we use a four digit
+Neat! So how and why does it work? For simplicity's sake, we use a fourdigit
number
.I abcd ,
where
@@ 685,9 +679,9 @@ represents the number of thousands,
.I b
the number of hundreds,
.I c
the number of tens and
+the number of tens, and the number
.I d
the number of ones. This number can be represented as
+This number can be represented as
.I a "" 1000
+
.I b "" 100
@@ 725,8 +719,8 @@ hi{ 11 left ( 91a ~+~ 9b ~+~ c right ) } ~~ a ~+~ b ~~ c ~+~ d
.EN
We can see that the first term in the expression is divisible by 11. This means
that if, and only if, the sum of the other terms is divisible by 11 the whole
expression is divisible by 11, and so the theorem is proven.
+that if, and only if, the sum of the other terms is divisible by 11, then the
+whole expression is divisible by 11, and so the theorem is proven.
.DIVS theorem
.EQ
@@ 753,13 +747,13 @@ The fraction flip when dividing
Many of you have probably been taught the trick of flipping the right fraction
in a division to instead use simpler multiplication. This is how it works:
We start of with the division:
+We start with the division:
.EQ
{ 3 over 2 } / { 4 over 5 }
.EN
From there we can reconstruct the two fractions as the dividend over
+From there, we can reconstruct the two fractions as the dividend over
the divisor with a horizontal line, for simplicity's sake, like so:
.EQ
@@ 767,10 +761,10 @@ the divisor with a horizontal line, for simplicity's sake, like so:
~=~ {{ 3 over 2 } over { 4 over 5 }}
.EN
After that the "trick" can begin. First we multiply both the dividend and the
+After that, the "trick" can begin. First, we multiply both the dividend and the
divisor with the inverse of the divisor. As long as we treat the dividend and
the divisor the same way, this is fine. Be aware of PEMDAS though! If any of
the dividend or the divisor would have been an addition or subtraction you
+the divisor the same way, this is fine. However, keep PEMDAS in mind! If any of
+the dividend or the divisor would have been an addition or subtraction, you
would have to multiply both terms by
.I x ,
either by
@@ 788,9 +782,9 @@ or
~=~ { 3 over 2 } ~times~ { 5 over 4 }
.EN
As you can see the right fraction has now "flipped", and not by magic, but with
logic and reason. As division by 1 is equal to the dividend, we can then solve
the expression, like so:
+As you can see, the right fraction has now "flipped", and not by magic, but
+with logic and reason. So, as division by 1 is equal to the dividend, we can
+then solve the expression, like so:
.EQ
{ 3 over 2 } ~times~ { 5 over 4 }
@@ 798,8 +792,8 @@ the expression, like so:
~=~ { 15 over 18 }
.EN
As the final cherry on top, we can prove the procedure by dividing 1 by 2 as we
know this should result in one half.
+As the final cherry on top, we can prove the procedure by dividing 1 by 2, as
+we know this should result in one half.
.EQ
{ 1 / 2 }
diff git a/noxz.tech/guides/index.www b/noxz.tech/guides/index.www
@@ 3,16 +3,6 @@ have acquired over the years \(en for various programs and concepts. I will
fill this list over a period of time, so it's far from complete \(en and
probably never will be.
.HnS 1
Concepts
.HnE

.DLS
.LIURL mathematics "Mathematics"
as for many hackers, is an interesting
and fascinating topic. It's literally a language to explain everything.
.DLE

.HnS 1
Programs
.HnE
diff git a/noxz.tech/guides/mathematics/.metadata b/noxz.tech/guides/mathematics/.metadata
@@ 1 +0,0 @@
.ds TITLE Mathematics