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How would I prove that for all elements of a set {n | n ∈ ℕ ∧ n > 7} (all natural numbers greater than 7), there are multiples of 3 and 5 which, when added up, are equal to that element of the set?

A := {n | n ∈ ℕ ∧ n > 7}
∀ n ∈ A: n = x*3 + y*5

I do understand why this is true: 

8 = 1*3 + 1*5
9 = 3*3 + 0*5
... 
15 = 0*3 + 3*5

and this pattern repeats once 16 is reached, as 8 can be replaced with 1*3 + 1*5, so I'd basically have to add 1 to x and y for every time 8 fits in n.

I am however not sure how I can formulate this mathematically correct.

Thanks in advance, CrushedPixel

Empy2
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CrushedPixel
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2 Answers2

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Without induction!

Let $q$ and $r$ be the quotient and remainder when dividing $n$ by $3$, so $n=3q+r$.

Case $r=0$: Then, of course, $n=q\cdot 3+0\cdot 5$.

Case $r=1$: Since $n=3q+1>7$ we must have $3q>6$ and so $q\ge 3$. Then we have $$ n = 3q+1 = 3(q-3)+9+1 = (q-3)\cdot3 + 2\cdot 5 $$

Case $r=2$: Since $n=3q+2>7$ we must have $3q>5$ and so $q\ge 2$. Then we have $$ n = 3q+2 = 3(q-1)+3+2 = (q-1)\cdot3 + 1\cdot 5$$

hmakholm left over Monica
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  • Thank you for this simple proof! While the induction was the more obvious solution (at least to me), this one is a lot simpler. – CrushedPixel Nov 18 '15 at 09:12
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Let's prove it by induction.

For $n=8$, we have $8=1 \cdot 3 + 1 \cdot5$.

Let $n>8$. By induction, $n-1= u\cdot 3+v\cdot 5$.

Write $v=3q+r$ with $r \in \{0,1,2\}$. Then $n-1=u'\cdot 3 + r \cdot 5$ with $u'=u+5q$.

If $r=0$, then $7 < n-1 = u' \cdot 3$ implies $u'\ge 3$. Then $n=(u'-3)\cdot 3 + 2 \cdot 5$.

If $r=1$, then $n=u' \cdot 3 + 6 = (u'+2)\cdot 3 + 0 \cdot 5$.

If $r=2$, then $n=u' \cdot 3 + 11 = (u'+2)\cdot 3 + 1 \cdot 5$.

lhf
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