caledonia Posted May 4, 2018 Share Posted May 4, 2018 I seek a proof, (using the properties of the h.c.f.), that if k divides nr then k = k_{1}k_{2} where k_{1} divides n and k_{2 }divides r Link to comment Share on other sites More sharing options...

mathematic Posted May 4, 2018 Share Posted May 4, 2018 Use fndamental theorem of arithmetic. https://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic Link to comment Share on other sites More sharing options...

Brett Nortj Posted August 2, 2018 Share Posted August 2, 2018 On 04/05/2018 at 8:39 PM, caledonia said: I seek a proof, (using the properties of the h.c.f.), that if k divides nr then k = k_{1}k_{2} where k_{1} divides n and k_{2 }divides r Well, as a layman, it would seem that if [k] = [k1k2], then [k] = [2] itself, as [1] * [2] = [2]. This is just speculation, let's get more intense? If the [k] = prime and divides into [k2], then six would be the lowest figure of the range, as the lowest figure for [k1] would be [3], as the lowest prime, so, we could run a log of [3] to find the 'ratios' of one to two, of course. this would follow, basically, that if [k1] = [3], then [k2] = [6], and k = [18], and that [k1] would actually be [k/3], and, that [k2] would be [k/2]. So, if [k=18] as [k=3=18] then [n] would be [k/3]. But, I am just a layman in this vast world. Link to comment Share on other sites More sharing options...

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