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From the book Discrete Mathematics for Computing 2nd Edition in eBook:

enter image description here

I know how to perform the Euclidean Algorithm and GCM(a,b). I am however, deeply confused by this:

$$1 = 415 - 69(421 - 1 \times 415)$$ $$ = 70 \times 415 - 69 \times 421$$

How is this expression constructed $70 \times 415 - 69 \times 421$?

N.B. I am still new to this and learning, I am using the following book(s):

Discrete Mathematics for Computing / Edition 2
by Peter Grossman [eBook]

Discrete Mathematics for Computing / Edition 3
by Peter Grossman [Physical copy]
Cornman
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Alix Blaine
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    Count how many $415$s are in that expression. (Answer: $1+69$.) – Randall Mar 13 '24 at 18:13
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    $415-69(-1\times415)=(1-(-69))(415)=70\times415$ – J. W. Tanner Mar 13 '24 at 18:14
  • @J.W.Tanner Thanks. Care to explain? This looks very confusing to me still. **Still confused since I do not see the steps from $1 = 415 - 69(421 - 1 \times 415)$ to $=70 \times 415 - 69 \times 421$? – Alix Blaine Mar 13 '24 at 18:15
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    It's just algebra: $x + ax= (1+a)x$. – Randall Mar 13 '24 at 18:19
  • @Randall, sure, but this whole expression jumping looks confusing, since we do not solve for anything specific i.e., freeing a variable. – Alix Blaine Mar 13 '24 at 18:21
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    You combine the $415$ terms and separate them from the $421$ term – J. W. Tanner Mar 13 '24 at 18:24
  • @J.W.Tanner you mean 421 - 415? Or, like similar take out each other? – Alix Blaine Mar 13 '24 at 18:25
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    You're just collecting like terms, but the "like terms" are $415$s and $421$s. – Randall Mar 13 '24 at 18:26
  • @Randall you mean similar terms negate each other, right? – Alix Blaine Mar 13 '24 at 18:27
  • @J.W.Tanner, so that means, 415 + 415 because there are two terms. – Alix Blaine Mar 13 '24 at 18:33
  • one of the "$415$" terms is multiplied by $(-69)(-1)$ – J. W. Tanner Mar 13 '24 at 18:33
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    By the way, there are a couple more worked examples over at the page on Bézout's identity on Brilliant if you want to see this arithmetic in action. – Sammy Black Mar 13 '24 at 18:35
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    Imagine if it was $x-69(y-x)$. Can you see how that equals $70x-69y$? – Malady Mar 13 '24 at 19:47
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    You do have $1=a-b(c-1\times a)=(b+1)a-bc$ – Ataulfo Mar 13 '24 at 20:03
  • @Malady Yes, in this context I can, because it doesn't look like the above. – Alix Blaine Mar 13 '24 at 20:18
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    Since you understand collecting like terms in $x-69(y-x) = 70x-69y$, will you find it more understandable to name all the dividends, divisors and remainders from the beginning? $r_0 = 2093$, $r_1 = 836$, $r_2=421$, ... Then the equations become: $$\begin{align} r_4 &= r_2 - 1r_3 \tag 3\ r_5 &= r_3 - 69r_4 \tag4\ &= r_3 - 69(r_2-1r_3)\ &= 70r_3-69r_2 \end{align}$$ – peterwhy Mar 13 '24 at 21:27
  • @AlixBlaine Okay, that equality is true for all values of $x,y$. Thus it is true when $x=415$, $y=421$. Does that make sense? – Malady Mar 14 '24 at 01:04
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    We can just do the following to get to the $70\times415-69\times421$: $$415-69(421-1\times415)=415-69(421-415)$$ Now don't calculate the $421-415$ just factorize. $$415-69(421-415)=415+(-69\times421)+(-69\times -415)=415-69\times421+69\times415=415+69\times415-69\times421$$

    Now we have $415+69\times415-69\times421$, in this expression we have one $415$ plus $69$ $415$s. That makes $70$ $415$s. So we have $70\times 415-69\times 421$.

    Finally we have $$415-69(421-1\times415)=70\times415-69\times421$$

     – David Krell Mar 16 '24 at 12:16
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    Wow. A fifty point bounty for a canonical explanation for a line or two of elementary high school algebra. That should attract lots of identical answers. – Ethan Bolker Mar 16 '24 at 15:47

3 Answers3

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How is this expression constructed $70 \times 415 - 69 \times 421$?

Using colors might be helpful.

$$\begin{align}1& = \color{red}{415} - 69(\color{blue}{421} - 1 \times \color{red}{415}) \\\\&=\color{red}{415} - 69\times \color{blue}{421} +69 \times\color{red}{415} \\\\&=\color{red}{415}+69 \times \color{red}{415} - 69\times \color{blue}{421} \\\\&=(1+69)\times \color{red}{415}- 69\times \color{blue}{421} \\\\&=70 \times \color{red}{415} - 69 \times \color{blue}{421}\end{align}$$

To get a solution $(x,y)$ of the equation $\color{purple}{2093}x+\color{orange}{836}y=1$, they started with $$\color{blue}{421}=\color{purple}{2093}-2\times \color{orange}{836}\tag1$$ $$\color{red}{415}=\color{orange}{836}-1\times \color{blue}{421}\tag2$$ $$\color{green}6=\color{blue}{421}-1\times \color{red}{415}\tag3$$ $$1=\color{red}{415}-69\times \color{green}6\tag4$$

They have $$\begin{align}1&=\color{red}{415}-69\times \color{green}6 \\\\&=\color{red}{415}-69\times (\color{blue}{421}-1\times \color{red}{415}) \\\\&=70 \times \color{red}{415} - 69 \times \color{blue}{421} \\\\&=70 \times (\color{orange}{836}-1\times \color{blue}{421})- 69 \times \color{blue}{421} \\\\&=70\times \color{orange}{836}-139\times \color{blue}{421} \\\\&=70\times \color{orange}{836}-139\times (\color{purple}{2093}-2\times \color{orange}{836}) \\\\&=\color{purple}{2093}\times (-139)+\color{orange}{836}\times 348\end{align}$$

Therefore, $(x,y)=(-139,348)$ is a solution of the equation $\color{purple}{2093}x+\color{orange}{836}y=1$.

mathlove
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  • How do you go from: $= 415 - 69 \times 421 + 69 \times 415$ to $= 415 + 69 \times 415 - 69 \times 421$ and from this to: $= (1 + 69) \times 415 - 69 \times 421$ ? I don't understand the steps and transitions that's happening there! – Alix Blaine Mar 17 '24 at 07:32
  • @Alix Blaine : Since we have that $A-B+C$ is equal to $A+C-B$, setting $A=415,B=69\times 421$ and $C=69\times 415$ gives $$415−69×421+69×415$$$$=415+69×415−69×421$$ Next, since we have that $E\times D+F\times D-G$ is equal to $(E+F)\times D-G$, setting $D=415,E=1,F=69$ and $G=69\times 421$ gives $$415+69×415−69×421$$$$=1\times 415+69\times 415-69\times 421$$$$=(1+69)\times 415-69\times 421$$ – mathlove Mar 17 '24 at 08:04
  • So you are doing some re-writing? – Alix Blaine Mar 17 '24 at 09:35
  • Any link where I can learn about this? – Alix Blaine Mar 17 '24 at 09:36
  • How to you arrive here: $1 \times 415 + 69 \times 415 - 69 \times 421$, I fail to see what happens exactly? – Alix Blaine Mar 17 '24 at 09:42
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    @Alix Blaine : "you are doing some re-writing?" Yes, I am. "Any link where I can learn about this?" You might want to see this. "How to you arrive here: $1×415+69×415−69×421$" Since we have that $A+B−C$ is equal to $1×A+B−C$ (just rewriting $A$ as $1\times A$), setting $A=415,B=69×415$ and $C=69×421$, we have $$415+69×415−69×421$$ $$=1×415+69×415−69×421$$ – mathlove Mar 17 '24 at 14:27
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You already brought up the Euclidean algorithm. For $n,m\in\mathbb{Z}$ with $\operatorname{gcd}(n,m)=1$ you get from "reversing" the algorithm an expression of $1=an+bm$.

Example:

$\operatorname{gcd}(13,23)=1$.

We have

$23=1\cdot 13+10$

$13=1\cdot 10+3$

$10=3\cdot 3+1$

Now use the last equation to get an expression in $1$ and substitute into the other equations to express this in terms of $23$ and $13$.

So $1=10-3\cdot 3$.

Now $13=1\cdot 10+3\Leftrightarrow 3=13-10$ (we solve for the "remainder" for the substitution)

Now plug this in for $3$.

$1=10-3\cdot (13-10)=10-3\cdot 13+3\cdot 10=4\cdot 10-3\cdot 13$.

Proceed and solve the first equation for $10$.

$10=23-13$.

Plug this in:

$1=4(23-13)-3\cdot 13=4\cdot 23-4\cdot 13-3\cdot 13=4\cdot 23-7\cdot 13$.

Cornman
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As far as I see, the calculations from the quotation concern to construct an integer solution $(x,y)$ of the Diophantine equation $2093x+836y=1$. To do this, we apply first the Euclidean algorithm to find the greatest common divisor to numbers $r_1=2093$ and $r_2=836$ and then go backwards. Namely, the Euclidean algorithm yields:

$r_3=421=2093-2\times 836=r_1-2\times r_2$

$r_4=415=836-1\times 421=r_2-1\times r_3$

$r_5=6=421-1\times 415=r_3-1\times r_4$

$r_6=1=415-69\times 6=r_4-69\times r_5$.

Next we go backwards:

$1=r_6=r_4-69\times r_5=415-69\times 6$

$1=r_4-69\times r_5=r_4-69\times (r_3-1\times r_4)=-69\times r_3+70\times r_4=-69\times 421+70\times 415$

$1=-69\times r_3+70\times r_4=-69\times r_3+70\times (r_2-1\times r_3)=70\times r_2-139\times r_3=70\times 836-139\times 421$

$1=70\times r_2-139\times r_3=70\times r_2-139\times (r_1-2\times r_2)=-139\times r_1+348\times r_2=-139\times 2093+348\times 836$

That is we constructed an integer solution $(x,y)=(-139,348)$ of the equation $2093x+836y=1$.

Alex Ravsky
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  • Hi, I am super confused by $r_n$ in your examples. – Alix Blaine Mar 16 '24 at 11:06
  • Can you simplify this with those numbers from the book example? – Alix Blaine Mar 16 '24 at 11:07
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    @AlixBlaine The $r_n$'s are consecutive remainders of the Euclidean algorithm. They are used to indicate what is going on. Yes, we can skip them, but then the calculations will become a distributivity exercise. :-) – Alex Ravsky Mar 16 '24 at 11:23
  • What do you mean by "a distributivity exercise"? – Alix Blaine Mar 16 '24 at 11:27
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    @AlixBlaine A sequence of arithmetic transformations, based on the distributive law, for instance: $1=415-69\times 6=415-69\times (421-415)=-69\times 421+70\times 415= -69\times 421+70\times (836-1\times 421)=70\times 836-139\times 421= 70\times 836-139\times (2093-2\times 836)=-139\times 2093+348\times 836.$

    I think, this does not explain a lot what is going on and from where we obtained the respective numbers.

     – Alex Ravsky Mar 16 '24 at 11:35