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I would like to verify whether my proof is correct. The answer sheet used a much more intuitive and logical approach but I think mine is correct also.

To prove: $7|13\cdot 6^{n}+8\cdot 13^{n}$ for all natural numbers

Proof: We proceed by induction and show the base case holds. $13+8=21 $. Since 7 divides 21 the base case holds.

We assume $7|13\cdot 6^{m}+8\cdot 13^{m}$ and we need to show $7|13\cdot6^{m+1}+8\cdot 13^{m+1}$.

Since, $7|13\cdot 6^{m}+8\cdot 13^{m}$, we have, $13\cdot 6^{m}+8\cdot 13^{m} =7x, \space x \in \mathbb{N}$.

We rewrite $13\cdot 6^{m+1}+8\cdot 13^{m+1}$ as $6(13\cdot 6^m)+13(8\cdot 13^m)$ and notice, $13\cdot 6^m=7x-8\cdot13^m$.

We substitute and find,

$6(7x-8\cdot13^m)+13(8\cdot13^m)=42x-48\cdot13^m+104\cdot13^m = 42x+56\cdot13^m=7(6x+8\cdot13^m)$

Since $6x+8\cdot13^m \space \in \mathbb{N}$

$\\ \therefore $ By the principle of mathematical induction, $7|13\cdot 6^{n}+8\cdot 13^{n}$ for all natural numbers $n \space \blacksquare$.

José Carlos Santos
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Cro-Magnon
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    Yes, seems correct. But note that $6\equiv -1\equiv 13$ and $8\equiv 1\pmod7$, so it's $\equiv (-1)\cdot(-1)^n+1\cdot(-1)^n,\pmod7$. – Berci Oct 09 '18 at 21:06
  • @Berci My answer sheet used that approach. So, $7 \cdot 13 \cdot 6^m - 13 \cdot 6^m $ etc. That's what you mean by congruency right? – Cro-Magnon Oct 09 '18 at 21:09
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    Must the proof use induction? Do you know modular arithmetic or congruences and can you use them? – Bill Dubuque Oct 09 '18 at 21:10
  • @BillDubuque I had to use induction but if you have time could you please answer this question how you would have done it? – Cro-Magnon Oct 09 '18 at 21:11
  • @Berci for some reason I found it hard to reason about it that way because of the combination of the multiplication of numbers. – Cro-Magnon Oct 09 '18 at 21:12
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    Just verify yourself that $a_1\equiv a_2,\ b_1\equiv b_2\implies a_1b_1\equiv a_2b_2\pmod m$ for arbitrary integers $a_i,b_i,m$. [$m,|,(a_2-a_1)b_2 + a_1(b_2-b_1)$] Everything else follows from here. (E.g. by induction, $a_1^k\equiv a_2^k \pmod m$ then, etc.) – Berci Oct 09 '18 at 21:17
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    I'd say. $!\bmod 7!:\ \color{#c00}{13}\equiv 6,\Rightarrow, 13\cdot 6^n + 8\cdot \color{#c00}{13}^n\equiv 21\cdot 6^n\equiv 0\cdot 6^n.\ $ In this answer I carefully explain the realtionship between the congruence and inductive proofs. – Bill Dubuque Oct 09 '18 at 21:23
  • @BillDubuque Thank you! – Cro-Magnon Oct 09 '18 at 21:27
  • @Berci Thank you! – Cro-Magnon Oct 09 '18 at 21:27
  • @Cro-Magnon I added an answer that should prove enlightening. – Bill Dubuque Oct 09 '18 at 22:10

4 Answers4

3

It is correct, but:

  • When you wrote that “We rewrite $13\cdot 6^{m}+8\cdot 13^{m}$ as $6(13\cdot 6^m)+13(8\cdot 13^m)$”, what you meant was “We rewrite $13\cdot 6^{m+1}+8\cdot 13^{m+1}$ as $6(13\cdot 6^m)+13(8\cdot 13^m)$”.
  • You don't need to compute $6\times8$ and $13\times8$. Note that\begin{align}6(7x-8\times13^m)+13(8\times13^m)&=7\times(6x)+(-6+13)\times(8\times13^m)\\&=7\times(6x+8\times13^m).\end{align}
José Carlos Santos
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  • Thank you! Your second point is really important but I found it hard to see and in fact didn't see it in my proof. – Cro-Magnon Oct 09 '18 at 21:39
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Per your comment, we simplify your inductive proof to highlight the relationship with congruence-based proofs. Your proof can be viewed as scaling by $\color{#c00}{13 = 6+7}\,$ the hypothesized equation $\,P(n)$

$$\begin{align} (\color{#c00}{13=6+7})\,\ {\rm times}\,\ (\overbrace{a\,13^{\large n}\ +\ \ b\,6^{\large n}\ \!=\ 7k}^{\Large P(n)})\qquad\quad&\\[.3em] \Longrightarrow\ \ \ \underbrace{a\,13^{\large n+1} + b\,6^{\large n+1}\ =\,\ 7(13k-b\,6^n)}_{\Large P(n+1)} \end{align}\qquad\qquad$$

Its arithmetical essence is clearer $\bmod 7\,$ using congruences and the Congruence Product Rule

$$\begin{align} \color{#c00}{13}\,&\equiv\, \color{#c00}6\\[.3em] \times\ \ \quad a\,13^{\large n}&\equiv -b\,6^{\large n}\quad \qquad P(n)\\[1em] \hline \Rightarrow\ a\,13^{\large n+1}&\equiv -b\,6^{\large n+1}\qquad P(n+1) \end{align}\quad \qquad$$

where $\,a,b = 8,13$ in your case. In this answer I explain at length how the inductive proofs are nothing but special cases of the proofs of the Congruence Product Rule. As above, it is much clearer to use the rule as a lemma (call by name) rather the repeat its proof inline (call by value) in the special case at hand.

Bill Dubuque
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$$13\cdot 6^{n}+8\cdot 13^{n}\equiv (-1)(-1)^n+(1)(-1)^n =0 \text { mod (7) }$$

Mohammad Riazi-Kermani
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Rewrite $13\cdot 6^n+8\cdot 13^n$ in base $7$: $$16\cdot 6^n+11\cdot 16^n$$

Note that $6\cdot 6=51$ in base $7$.

The first term ends with $1$ if $n$ is even and ends with $6$ if $n$ is odd. Conversely, the second term ends with $1$ if $n$ is even and ends with $6$ if $n$ is odd. Therefore, the sum ends with $0$.

ajotatxe
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