prime number in action
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- Posts:41
- Joined:Tue Dec 21, 2010 10:17 pm
prove that for every integer k the numbers 2k+1 and 9k+4 are relatively prime.
and find G.C.D of the two numbers (2k-1 and 9k+4) in terms of k or as a function of k.
the arrivals
and find G.C.D of the two numbers (2k-1 and 9k+4) in terms of k or as a function of k.
the arrivals
women of purity are for men of purity and hence men of purity are for women of purity - THE HOLY QURAN
Re: prime number in action
Part 1: $9(2k+1)-2(9k+4)=1$. So, if some $d$ divides both $2k+1$ and $9k+4$, then it divides $1$.
So the gcd is $1$.
Part 2: $2(9k+4)-9(2k-1)=17$. So the gcd is $1 \text{ or } 17$.
(It is easy to check that both we can attain both values as gcds)
So the gcd is $1$.
Part 2: $2(9k+4)-9(2k-1)=17$. So the gcd is $1 \text{ or } 17$.
(It is easy to check that both we can attain both values as gcds)
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
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Re: prime number in action
how will you approach this
(a-b,a+b) greater or equal (a,b)
(a,b) means their gcd function
(a-b,a+b) greater or equal (a,b)
(a,b) means their gcd function
women of purity are for men of purity and hence men of purity are for women of purity - THE HOLY QURAN
Re: prime number in action
$(a-b,a+b)=(a+b+a-b,a+b-a+b)=(2a,2b)=2(a,b) \geq (a,b)$
Re: prime number in action
Actually it is incorrect. For example, take $a=7,=4$. Then $(a,b)=(a-b,a+b)=1$abir91 wrote:$(a-b,a+b)=(a+b+a-b,a+b-a+b)=(2a,2b)=2(a,b) \geq (a,b)$
Solution:
Let $(a,b)=d$. Then $(a,b)=(da',db')=d(a',b')=d$
Now $(a-b,a+b)=d(a'-b',a'+b')$. We have $(a'+b)+(a'-b')=2a'$. So, $(a-b,a+b)=d \text { or } 2d \geq (a,b)$
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
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- leonardo shawon
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Re: prime number in action
amm, sorry, but whats GCD ?
and i didnt understand ur last two lines moon bhaia., and amm did u take a=7 and b=4 ??
and i didnt understand ur last two lines moon bhaia., and amm did u take a=7 and b=4 ??
Ibtehaz Shawon
BRAC University.
long way to go .....
BRAC University.
long way to go .....
Re: prime number in action
GCD=Greatest Common Divisor (গসাগু in Bengali). For example the gcd of $2,4$ is $2$.
In the last two lines at first I showed that $(a-b,a+b)=d(a'-b-,a'+b')$. Then I examined the gcd of $(a'-b,a'+b')$.
The gcd must divide their sum=$2a'$. So it must divide $2$. So the gcd is either $2$ or $1$. (and therefore $(a+b,a-b)=2d, d$.)
BTW $7,4$ was a special example. Here we are taking about any $a,b$.
In the last two lines at first I showed that $(a-b,a+b)=d(a'-b-,a'+b')$. Then I examined the gcd of $(a'-b,a'+b')$.
The gcd must divide their sum=$2a'$. So it must divide $2$. So the gcd is either $2$ or $1$. (and therefore $(a+b,a-b)=2d, d$.)
BTW $7,4$ was a special example. Here we are taking about any $a,b$.
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
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Re: prime number in action
আরে তাইতো, কি লেখছি! যাই হোক, তুমি না লিখা অন্যদের দিতে পারতা কোথায় ভুল হইছে।
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Re: prime number in action
but picking up your example a=7,b=4 it does fulfil the require statement of the given condition of the question.whers the fault???
women of purity are for men of purity and hence men of purity are for women of purity - THE HOLY QURAN
Re: prime number in action
Let $g=gcd(2k-1,9k+4)$,then $g|2k-1|18k-9,18k+8$ so their difference too.So $g|17,g=1$ or $17$the arrivals wrote:prove that for every integer k the numbers 2k+1 and 9k+4 are relatively prime.
and find G.C.D of the two numbers (2k-1 and 9k+4) in terms of k or as a function of k.
the arrivals
First part is similar
One one thing is neutral in the universe, that is $0$.