Prove that $10^n \equiv 1 \pmod 3$ for all positive integers $n$ by mathematical induction.
Can someone please help me in solving this problem and explain what's going on? Any guidance would be greatly appreciated.
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Daniel W. Farlow
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Roger
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base case $n = 1 \Rightarrow 10 \equiv 1\mod 3$ holds. Assume that holds for $n = k$. Then for $n = k+1$, we have: $$10^{k+1} \mod 3 = (10\mod 3)*(10^{k}\mod 3) = (10^k \mod3) \equiv 1\mod 3$$ notice that $10^k$ holds by the induction hypothesis.
okie
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That's it to the problem? – Roger Apr 17 '15 at 01:25
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@Roger that is a simplified answer, notice that the multiplicative property holds in modular arithmetic, namely: $(a \mod n)(b \mod n) = ab \mod n$. Do you understand the principle of mathematical induction? – okie Apr 17 '15 at 01:27
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Yes, I do. I just want to be certain that what's written out above satisfies what the question is asking for. – Roger Apr 17 '15 at 01:28
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sure, since the assumption hypothesis holds for $n = k$ and since: $10 \mod 3 \equiv 1$, then solution is true – okie Apr 17 '15 at 01:30
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Thank you my friend. Have a check! – Roger Apr 17 '15 at 01:33
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You are mixing up the operator mod with the congruence. The LHS equality is not true as written. – Bill Dubuque Apr 17 '15 at 01:36
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@BillDubuque opps my bad, if you can fix such answer i will appreciate that. – okie Apr 17 '15 at 01:42
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@Roger $,{\rm mod}\ 3!:\ \color{#c00}{10\equiv 1},\Rightarrow, 10^{k+1}!\equiv\color{#c00}{10}\cdot 10^k \equiv \color{#c00}{1}\cdot \color{#0a0}{10^k}\equiv \color{#c00}1\cdot \color{#0a0}1 \equiv 1\ $ by $,\rm\color{#0a0}{induction},$ and Congruence Product Rule – Bill Dubuque Apr 17 '15 at 02:04
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Here is perhaps a more straightforward inductive proof: Let $S(n)$ denote the claim $$ S(n) : 10^n\equiv 1\pmod{3}\Longleftrightarrow 10^n-1=3m, m\in\mathbb{Z}. $$ Base case ($n=1$): $S(1)$ says that $10^1-1=3m, m\in\mathbb{Z}$, and we know this is true because $10^1-1=9=3(3)$.
Inductive step: Fix some $k\geq 1$ and assume that $S(k)$ is true where $$ S(k) : 10^k\equiv 1\pmod{3}\Longleftrightarrow 10^k-1=3\ell, \ell\in\mathbb{Z}. $$ To be shown is that $S(k+1)$ follows where $$ S(k+1) : 10^{k+1}\equiv 1\pmod{3}\Longleftrightarrow 10^{k+1}-1=3\eta, \eta\in\mathbb{Z}. $$ Beginning with the left-hand side of $S(k+1)$, \begin{align} 10^{k+1} &= 10^k\cdot 10\tag{by definition}\\[0.5em] &= (3\ell +1)\cdot 10\tag{by $S(k)$, the ind. hyp.}\\[0.5em] &= 30\ell + 10\tag{expand}\\[0.5em] &= 3(10\ell+3)+1\tag{rearrange}\\[0.5em] &= 3\eta+1, \tag{$\eta=10\ell+3; \eta\in\mathbb{Z}$} \end{align} we end up at the right-hand side of $S(k+1)$, thus completing the inductive step.
By mathematical induction, $S(n)$ is true for all $n\geq 1$. $\blacksquare$
Daniel W. Farlow
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