Induction proof on a linear function

Asked Apr 23 '24 at 16:36

Active Apr 23 '24 at 17:53

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Am I on the right track with my attempt at the following problem?

Show that for all $n \in \mathbb{N}$ with $n \geq 8$ there exist $k, l \in \mathbb{N}$ such that $n=3 k+5 l$.

Hint: Use not only $n=8$ for the induction base.

Base case

$n = 8:$
$3\cdot1 + 5\cdot1 = 8$

$n + 1 = 9:$
$3\cdot1 + 5\cdot 1 + 1 = 9$

Induction step

Assume $n = 3k + 5l$ and $n+1 = 3k' + 5l'$

Claim: for $n + 2$ there exist some $k''$ and $l''$ such that $3k'' + 5l'' = n+2$

$3(k'-k) + 5(l'-l) -1 = n$

$3(k'-k) + 5(l'-l) +1 = n + 2$

Thank you in advance

edited Apr 23 '24 at 17:53

FD_bfa

4,757

asked Apr 23 '24 at 16:36

K-1

Does this answer your question? Proof that $n = 3k + 5l$ for $n > 7$ - found using an Approach0 search. FYI, there are other duplicates, e.g., Strong Induction: Every natural number $n\geq 8$ can be represented as $n=3k + 5\ell$, and other closely related questions involving the Frobenius Coin problem. – John Omielan Apr 23 '24 at 17:11
You haven't solve the base case $n=9.$ – Thomas Andrews Apr 23 '24 at 17:32
1

@JohnOmielan Yes, it does. I wasn't aware of Approach0 search . Thanks for introducing it here. Next time I try searching there before posting. – K-1 Apr 24 '24 at 20:08

Induction proof on a linear function

Base case

Induction step

0 Answers0