Question

I don't understand this question. Could someone help me understand it?

"Show that if \(a\) and \(b\) are positive integers, then there is a smallest positive ingeter of the form \(a-bk\), \(k\in\mathbb{N}\)."\[\]

Answer 1

Ask satellite

Answer 2

lol

Answer 3

I don't know what the question is asking for.

Answer 4

not sure either off the top of my head, so lets just pick something and see what it means

Answer 5

pick a = 10, b = 4

Answer 6

By the way, I made a typo: at the end, it should be: \(k\in\mathbb{Z}\).\[\]

Answer 7

then
\[10-6\]
\[10-2\times 4=2\] and \[10-3\times 4=-2\] so in this case k = 2

Answer 8

i mean i think intuitively it is pretty obvious. you need
\[a-kb>0\] and also an integer

Answer 9

i.e. a natural number

Answer 10

I knew there was something missing. The greater than zero part.

Answer 11

Should I break it down into three cases then? That is, when \(a<b\), \(a>b\) and \(a=b\)?\[\]

Answer 12

i don't know i had anesthesia and you are making me think. give me a moment or call in the big guns

Answer 13

ok lets try this. the hint is "prove there is a smallest one" and that means you are headed for "well ordering principle"

Answer 14

That's correct.

Answer 15

that says every set of non - negative integers has a least element (non-empty set)

Answer 16

so i am thinking that your real job is to define the set
\[M=\{a-bk:a-bk>0\}\] and show that is it not empty, then you will be done, in other words you just have to find some k (since a and b are given) that satisfies this inequality

Answer 17

Wow, that makes perfect sense. I'll try that and see what I can come up with. Thank you, sir!

Answer 18

yw and although i don't want to give away the farm, don't forget that a and b are POSITIVE integers, so don't look too hard for k!

Answer 19

Would this work for \(k\)?\[k=\left \lfloor \frac{a}{b}-\frac{1}{2} \right \rfloor\]

Answer 20

Hi pre-algebra!

Answer 21

Hi pre-algebra!

Answer 22

I did over-complicate the problem. So I re-arranged my thoughts and, like you said, chose to prove\[M=\left \{ (a-bk)\in\mathbb{N}:k\in\mathbb{Z} \right \}\]is non-empty.

I was so silly that I overlooked the simple fact that if \(k\leq0\), then \((a-bk)\in\mathbb{N}\).

-.-

Answer 23

since
\[a>0\] you can show the set is not empty by taking
\[k=0\]