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So, according to the book, for all $a, b, c$ that are elements of integers, it holds that $a|b$ implies $a|bx$ for all $x$ that is an element of integers. In other words it works for all ARBITRARY $x$ in the universe $Z$.

However, please consider this question:

When $2a + 3b$ is a multiple of $17$, prove that $17$ divides $9a + 5b$.

Proof(textbook):

We observe that $17|(2a + 3b) \implies 17|(-4)(2a + 3b)$ by the theorem where $a|b$ implies $a|bx$ for all $x$ that is an element of integers. Also since $17|17$ it follows that $17|[(17a + 17b) + (-4)(2a + 3b)]$ and consequently this simplifies to $17|(9a + 5b)$.

My problem with proof:

The book chooses the $x = -4$ for the arbitrary $x$ that is part of the universe $Z$, but I find that this isn't arbitrary at all because I'm pretty sure that if I used any other number for $x$ in the universe of $Z$ it wouldn't work with the proof. It would then seem that the book specifically chose it as $-4$ because the part where they include $17|17$ seems to be specific towards $-4$ being $x$ as well. Am I right in this? How would I solve questions like these?

Bill Dubuque
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Belphegor
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3 Answers3

4

HINT:

Eliminate one unknown $$9(2a+3b)-2(9a+5b)=?$$

lab bhattacharjee
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The proof given is an instance of a magic proof. It works quite nicely, but gives little indication of how one reaches it.

Let us compare $2a+3b$ and $9a+5b$. Can we multiply the $2$ by something to get a result which is congruent to $9$ modulo $17$? Yes, multiplying by $13$ will do it. Let us see what this does to the whole expression $2a+3b$.

We get $$13(2a+3b)=26a+39b\equiv 9a+5b\pmod{17}.$$ Since $17$ divides $2a+3b$, it divides $13(2a+3b)$. But since this is congruent to $9a+5b$ modulo $17$, it follows that $9a+5b$ is divisible by $17$.

Note that $-4\equiv 13\pmod{17}$, so, modulo $17$, multiplying by $-4$ has the same effect as multiplying by $13$.

Now we can if we wish hide our preliminary work, and just say multiply by $13$, or, more mysteriously still, by $-4$.

André Nicolas
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  • What is this process for turning 2 into 9 exactly called? I searched up modular arithmetic and couldn't find anything about turning one numebr into another. – Belphegor Jul 24 '14 at 05:43
  • Ordinarily, it is thought of as a two step process: turn the $2$ into a $1$ by multiplying $2$ by its modular inverse (that is a key term to search) and then turn the $1$ into $9$ by multiplying by $9$. The modular inverse of $2$ is $9$, since $2\cdot 9=18$. Multiply by $9$ again, to get $81$, which is congruent to $13$. For primes $p$ uite a bit bigger than $17$, modular inverses are found by using the extended Euclidean algorithm. – André Nicolas Jul 24 '14 at 06:02
  • Bu for our numbers $9$ and $2$, you can think of it as dividing $9$ by $2$ modulo $17$. Of course you can't divide $9$ by $2$, but $9$ is congruent to $9+17$, which you can divide by $2$. – André Nicolas Jul 24 '14 at 06:05
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Hint: convert to congruences, then scale first by $\,\large \frac{\color{#0a0}9}{\color{#c00}2}\ $ so $\ \color{#c00}2a\,\to\, \color{#0a0}9a\, $ so they have same coef of $\,a$

$$\bmod 17\!:\,\ 0\equiv \color{#c00}2a\!+\!3b\iff 0\equiv { \frac{\color{#0a0}9}{\color{#c00}2}}(\color{#c00}{2}a\!+\!3b)\equiv \color{#0a0}{9}a\!+\!{\frac{10}2}\,b\qquad\qquad\quad\! $$

Recall that equations (congruences) are preserved by scalings by invertible numbers (here our invertible scale factor is $\,9/2 := 9\cdot 2^{-1} \pmod{\!17},\,$ cf. modular fractions).

Geometrically, if we change notation $\,a,b \to y,x\,$ and view the equations as lines, then we see they are equivalent because they are same slope lines through a common point (origin):

$\ \ \ \begin{array}{l} 2y+3x\equiv\, 0\iff y\,\equiv\, \color{#c00}{-\frac{3}2}x\\ 9y+5x\equiv\, 0\iff y\,\equiv\, \color{#0a0}{-\frac{5}9}x\end{array}\ \,$ have equal slopes $ \, \color{#c00}{-\dfrac{3}2}\equiv \color{#0a0}{-\dfrac{5}9}\ $ by $ \ 3\cdot 9\equiv 2\cdot 5\:\!\pmod{\!17}$

Bill Dubuque
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