Numerical example for $gcd(a,b)=prod p_i^min(a_i,b_i)$

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I'm actually having trouble understanding the above corollary. Can anyone please provide a numerical example of that corollary?
Thank You So Very Much in advance.
Corollary If $a=prod p_i^a_i$ and $b=prod p_i^b_i$ where the $p_i$ are distinct primes, then $gcd(a,b)=prod p_i^min(a_i,b_i)$. The LCM is $prod p_i^max(a_i,b_i)$ and the product of the last two expressions is $ab$.
elementary-number-theory greatest-common-divisor prime-factorization least-common-multiple
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add a comment |
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I'm actually having trouble understanding the above corollary. Can anyone please provide a numerical example of that corollary?
Thank You So Very Much in advance.
Corollary If $a=prod p_i^a_i$ and $b=prod p_i^b_i$ where the $p_i$ are distinct primes, then $gcd(a,b)=prod p_i^min(a_i,b_i)$. The LCM is $prod p_i^max(a_i,b_i)$ and the product of the last two expressions is $ab$.
elementary-number-theory greatest-common-divisor prime-factorization least-common-multiple
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add a comment |
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I'm actually having trouble understanding the above corollary. Can anyone please provide a numerical example of that corollary?
Thank You So Very Much in advance.
Corollary If $a=prod p_i^a_i$ and $b=prod p_i^b_i$ where the $p_i$ are distinct primes, then $gcd(a,b)=prod p_i^min(a_i,b_i)$. The LCM is $prod p_i^max(a_i,b_i)$ and the product of the last two expressions is $ab$.
elementary-number-theory greatest-common-divisor prime-factorization least-common-multiple
$endgroup$
I'm actually having trouble understanding the above corollary. Can anyone please provide a numerical example of that corollary?
Thank You So Very Much in advance.
Corollary If $a=prod p_i^a_i$ and $b=prod p_i^b_i$ where the $p_i$ are distinct primes, then $gcd(a,b)=prod p_i^min(a_i,b_i)$. The LCM is $prod p_i^max(a_i,b_i)$ and the product of the last two expressions is $ab$.
elementary-number-theory greatest-common-divisor prime-factorization least-common-multiple
elementary-number-theory greatest-common-divisor prime-factorization least-common-multiple
edited Feb 17 at 16:58
Martin Sleziak
44.8k10119273
44.8k10119273
asked Feb 17 at 14:20
Jessi JhaJessi Jha
233
233
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3 Answers
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Certainly: consider $a=48=2^4cdot 3$ and $b=40=2^3cdot 5$. Then we let $p_1=2, p_2=3, p_3=5$ and observe that $a_1=4, a_2=1, a_3=0$ and $b_1=3, b_2=0, b_3=1$. Your corollary then tells us that
beginalign*
gcd(48,40)
&=gcd(a,b) \
&= prod p_i^min(a_i,b_i) \
&= p_1^min(a_1,b_1)cdot p_2^min(a_2,b_2)cdot p_3^min(a_3,b_3) \
&= 2^min(4,3)cdot 3^min(1,0)cdot 5^min(0,1) \
&= 2^3cdot 3^0 cdot 5^0 \
&= 8,
endalign*
Using a similar method we determine that
beginalign*
lcm(48,40)
&=lcm(a,b) \
&= prod p_i^max(a_i,b_i) \
&= 2^4cdot 3^1 cdot 5^1 \
&= 240.
endalign*
We can then observe that $$ab=48cdot 40=1920=8cdot 140=gcd(48,40)cdot lcm(48,40)=gcd(a,b)cdot lcm(a,b).$$
Let me know if anything needs clarifying (and if anyone knows the command that Latex recognises as a math operator for lcm).
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Formula for LCM should use Max and not minumum. The calculations are right sl
$endgroup$
– Edcookie274
Feb 17 at 15:23
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Thanks for pointing that out. Fixed.
$endgroup$
– Ben
Feb 17 at 15:32
1
$begingroup$
Thank You so much!!!! Your answer gave me complete clarity! !!! Thank you once again!!!
$endgroup$
– Jessi Jha
Feb 20 at 3:18
add a comment |
$begingroup$
$$beginaligna=600&=2^3cdot 3^1 cdot 5^2\
b=54&=2^1cdot 3^3 cdot 5^0\[2ex]
textGCD(a,b)&=2^1cdot 3^1cdot 5^0=6\
textLCM(a,b)&=2^3cdot 3^3cdot 5^2=5400\[2ex]
textGCD(a,b)cdottextLCM(a,b)&=2^4cdot 3^4cdot 5^2=32400=ab
endalign$$
$endgroup$
$begingroup$
Thank You so much!!!! All these examples really helped me!!
$endgroup$
– Jessi Jha
Feb 20 at 3:19
add a comment |
$begingroup$
For example, $540 = 2^2cdot 3^3cdot 5$ and $72 = 2^3cdot 3^2$, so their GCD is $2^2cdot 3^2 = 36$ and their LCM is $2^3cdot 3^3cdot 5 = 1080$. The point is that if a number $d$ is to divide both $a$ and $b$, the power to which each prime appears in the factorization of $d$ cannot exceed the mininum power to which it appears in the factorization of both $a$ and $b$. A similar analysis works for the LCM.
$endgroup$
$begingroup$
Thank You so much!!! I got it! !
$endgroup$
– Jessi Jha
Feb 20 at 3:21
add a comment |
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3 Answers
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3 Answers
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$begingroup$
Certainly: consider $a=48=2^4cdot 3$ and $b=40=2^3cdot 5$. Then we let $p_1=2, p_2=3, p_3=5$ and observe that $a_1=4, a_2=1, a_3=0$ and $b_1=3, b_2=0, b_3=1$. Your corollary then tells us that
beginalign*
gcd(48,40)
&=gcd(a,b) \
&= prod p_i^min(a_i,b_i) \
&= p_1^min(a_1,b_1)cdot p_2^min(a_2,b_2)cdot p_3^min(a_3,b_3) \
&= 2^min(4,3)cdot 3^min(1,0)cdot 5^min(0,1) \
&= 2^3cdot 3^0 cdot 5^0 \
&= 8,
endalign*
Using a similar method we determine that
beginalign*
lcm(48,40)
&=lcm(a,b) \
&= prod p_i^max(a_i,b_i) \
&= 2^4cdot 3^1 cdot 5^1 \
&= 240.
endalign*
We can then observe that $$ab=48cdot 40=1920=8cdot 140=gcd(48,40)cdot lcm(48,40)=gcd(a,b)cdot lcm(a,b).$$
Let me know if anything needs clarifying (and if anyone knows the command that Latex recognises as a math operator for lcm).
$endgroup$
$begingroup$
Formula for LCM should use Max and not minumum. The calculations are right sl
$endgroup$
– Edcookie274
Feb 17 at 15:23
$begingroup$
Thanks for pointing that out. Fixed.
$endgroup$
– Ben
Feb 17 at 15:32
1
$begingroup$
Thank You so much!!!! Your answer gave me complete clarity! !!! Thank you once again!!!
$endgroup$
– Jessi Jha
Feb 20 at 3:18
add a comment |
$begingroup$
Certainly: consider $a=48=2^4cdot 3$ and $b=40=2^3cdot 5$. Then we let $p_1=2, p_2=3, p_3=5$ and observe that $a_1=4, a_2=1, a_3=0$ and $b_1=3, b_2=0, b_3=1$. Your corollary then tells us that
beginalign*
gcd(48,40)
&=gcd(a,b) \
&= prod p_i^min(a_i,b_i) \
&= p_1^min(a_1,b_1)cdot p_2^min(a_2,b_2)cdot p_3^min(a_3,b_3) \
&= 2^min(4,3)cdot 3^min(1,0)cdot 5^min(0,1) \
&= 2^3cdot 3^0 cdot 5^0 \
&= 8,
endalign*
Using a similar method we determine that
beginalign*
lcm(48,40)
&=lcm(a,b) \
&= prod p_i^max(a_i,b_i) \
&= 2^4cdot 3^1 cdot 5^1 \
&= 240.
endalign*
We can then observe that $$ab=48cdot 40=1920=8cdot 140=gcd(48,40)cdot lcm(48,40)=gcd(a,b)cdot lcm(a,b).$$
Let me know if anything needs clarifying (and if anyone knows the command that Latex recognises as a math operator for lcm).
$endgroup$
$begingroup$
Formula for LCM should use Max and not minumum. The calculations are right sl
$endgroup$
– Edcookie274
Feb 17 at 15:23
$begingroup$
Thanks for pointing that out. Fixed.
$endgroup$
– Ben
Feb 17 at 15:32
1
$begingroup$
Thank You so much!!!! Your answer gave me complete clarity! !!! Thank you once again!!!
$endgroup$
– Jessi Jha
Feb 20 at 3:18
add a comment |
$begingroup$
Certainly: consider $a=48=2^4cdot 3$ and $b=40=2^3cdot 5$. Then we let $p_1=2, p_2=3, p_3=5$ and observe that $a_1=4, a_2=1, a_3=0$ and $b_1=3, b_2=0, b_3=1$. Your corollary then tells us that
beginalign*
gcd(48,40)
&=gcd(a,b) \
&= prod p_i^min(a_i,b_i) \
&= p_1^min(a_1,b_1)cdot p_2^min(a_2,b_2)cdot p_3^min(a_3,b_3) \
&= 2^min(4,3)cdot 3^min(1,0)cdot 5^min(0,1) \
&= 2^3cdot 3^0 cdot 5^0 \
&= 8,
endalign*
Using a similar method we determine that
beginalign*
lcm(48,40)
&=lcm(a,b) \
&= prod p_i^max(a_i,b_i) \
&= 2^4cdot 3^1 cdot 5^1 \
&= 240.
endalign*
We can then observe that $$ab=48cdot 40=1920=8cdot 140=gcd(48,40)cdot lcm(48,40)=gcd(a,b)cdot lcm(a,b).$$
Let me know if anything needs clarifying (and if anyone knows the command that Latex recognises as a math operator for lcm).
$endgroup$
Certainly: consider $a=48=2^4cdot 3$ and $b=40=2^3cdot 5$. Then we let $p_1=2, p_2=3, p_3=5$ and observe that $a_1=4, a_2=1, a_3=0$ and $b_1=3, b_2=0, b_3=1$. Your corollary then tells us that
beginalign*
gcd(48,40)
&=gcd(a,b) \
&= prod p_i^min(a_i,b_i) \
&= p_1^min(a_1,b_1)cdot p_2^min(a_2,b_2)cdot p_3^min(a_3,b_3) \
&= 2^min(4,3)cdot 3^min(1,0)cdot 5^min(0,1) \
&= 2^3cdot 3^0 cdot 5^0 \
&= 8,
endalign*
Using a similar method we determine that
beginalign*
lcm(48,40)
&=lcm(a,b) \
&= prod p_i^max(a_i,b_i) \
&= 2^4cdot 3^1 cdot 5^1 \
&= 240.
endalign*
We can then observe that $$ab=48cdot 40=1920=8cdot 140=gcd(48,40)cdot lcm(48,40)=gcd(a,b)cdot lcm(a,b).$$
Let me know if anything needs clarifying (and if anyone knows the command that Latex recognises as a math operator for lcm).
edited Feb 17 at 15:31
answered Feb 17 at 14:34
BenBen
329110
329110
$begingroup$
Formula for LCM should use Max and not minumum. The calculations are right sl
$endgroup$
– Edcookie274
Feb 17 at 15:23
$begingroup$
Thanks for pointing that out. Fixed.
$endgroup$
– Ben
Feb 17 at 15:32
1
$begingroup$
Thank You so much!!!! Your answer gave me complete clarity! !!! Thank you once again!!!
$endgroup$
– Jessi Jha
Feb 20 at 3:18
add a comment |
$begingroup$
Formula for LCM should use Max and not minumum. The calculations are right sl
$endgroup$
– Edcookie274
Feb 17 at 15:23
$begingroup$
Thanks for pointing that out. Fixed.
$endgroup$
– Ben
Feb 17 at 15:32
1
$begingroup$
Thank You so much!!!! Your answer gave me complete clarity! !!! Thank you once again!!!
$endgroup$
– Jessi Jha
Feb 20 at 3:18
$begingroup$
Formula for LCM should use Max and not minumum. The calculations are right sl
$endgroup$
– Edcookie274
Feb 17 at 15:23
$begingroup$
Formula for LCM should use Max and not minumum. The calculations are right sl
$endgroup$
– Edcookie274
Feb 17 at 15:23
$begingroup$
Thanks for pointing that out. Fixed.
$endgroup$
– Ben
Feb 17 at 15:32
$begingroup$
Thanks for pointing that out. Fixed.
$endgroup$
– Ben
Feb 17 at 15:32
1
1
$begingroup$
Thank You so much!!!! Your answer gave me complete clarity! !!! Thank you once again!!!
$endgroup$
– Jessi Jha
Feb 20 at 3:18
$begingroup$
Thank You so much!!!! Your answer gave me complete clarity! !!! Thank you once again!!!
$endgroup$
– Jessi Jha
Feb 20 at 3:18
add a comment |
$begingroup$
$$beginaligna=600&=2^3cdot 3^1 cdot 5^2\
b=54&=2^1cdot 3^3 cdot 5^0\[2ex]
textGCD(a,b)&=2^1cdot 3^1cdot 5^0=6\
textLCM(a,b)&=2^3cdot 3^3cdot 5^2=5400\[2ex]
textGCD(a,b)cdottextLCM(a,b)&=2^4cdot 3^4cdot 5^2=32400=ab
endalign$$
$endgroup$
$begingroup$
Thank You so much!!!! All these examples really helped me!!
$endgroup$
– Jessi Jha
Feb 20 at 3:19
add a comment |
$begingroup$
$$beginaligna=600&=2^3cdot 3^1 cdot 5^2\
b=54&=2^1cdot 3^3 cdot 5^0\[2ex]
textGCD(a,b)&=2^1cdot 3^1cdot 5^0=6\
textLCM(a,b)&=2^3cdot 3^3cdot 5^2=5400\[2ex]
textGCD(a,b)cdottextLCM(a,b)&=2^4cdot 3^4cdot 5^2=32400=ab
endalign$$
$endgroup$
$begingroup$
Thank You so much!!!! All these examples really helped me!!
$endgroup$
– Jessi Jha
Feb 20 at 3:19
add a comment |
$begingroup$
$$beginaligna=600&=2^3cdot 3^1 cdot 5^2\
b=54&=2^1cdot 3^3 cdot 5^0\[2ex]
textGCD(a,b)&=2^1cdot 3^1cdot 5^0=6\
textLCM(a,b)&=2^3cdot 3^3cdot 5^2=5400\[2ex]
textGCD(a,b)cdottextLCM(a,b)&=2^4cdot 3^4cdot 5^2=32400=ab
endalign$$
$endgroup$
$$beginaligna=600&=2^3cdot 3^1 cdot 5^2\
b=54&=2^1cdot 3^3 cdot 5^0\[2ex]
textGCD(a,b)&=2^1cdot 3^1cdot 5^0=6\
textLCM(a,b)&=2^3cdot 3^3cdot 5^2=5400\[2ex]
textGCD(a,b)cdottextLCM(a,b)&=2^4cdot 3^4cdot 5^2=32400=ab
endalign$$
edited Feb 17 at 14:43
answered Feb 17 at 14:32
Daniel MathiasDaniel Mathias
1,36018
1,36018
$begingroup$
Thank You so much!!!! All these examples really helped me!!
$endgroup$
– Jessi Jha
Feb 20 at 3:19
add a comment |
$begingroup$
Thank You so much!!!! All these examples really helped me!!
$endgroup$
– Jessi Jha
Feb 20 at 3:19
$begingroup$
Thank You so much!!!! All these examples really helped me!!
$endgroup$
– Jessi Jha
Feb 20 at 3:19
$begingroup$
Thank You so much!!!! All these examples really helped me!!
$endgroup$
– Jessi Jha
Feb 20 at 3:19
add a comment |
$begingroup$
For example, $540 = 2^2cdot 3^3cdot 5$ and $72 = 2^3cdot 3^2$, so their GCD is $2^2cdot 3^2 = 36$ and their LCM is $2^3cdot 3^3cdot 5 = 1080$. The point is that if a number $d$ is to divide both $a$ and $b$, the power to which each prime appears in the factorization of $d$ cannot exceed the mininum power to which it appears in the factorization of both $a$ and $b$. A similar analysis works for the LCM.
$endgroup$
$begingroup$
Thank You so much!!! I got it! !
$endgroup$
– Jessi Jha
Feb 20 at 3:21
add a comment |
$begingroup$
For example, $540 = 2^2cdot 3^3cdot 5$ and $72 = 2^3cdot 3^2$, so their GCD is $2^2cdot 3^2 = 36$ and their LCM is $2^3cdot 3^3cdot 5 = 1080$. The point is that if a number $d$ is to divide both $a$ and $b$, the power to which each prime appears in the factorization of $d$ cannot exceed the mininum power to which it appears in the factorization of both $a$ and $b$. A similar analysis works for the LCM.
$endgroup$
$begingroup$
Thank You so much!!! I got it! !
$endgroup$
– Jessi Jha
Feb 20 at 3:21
add a comment |
$begingroup$
For example, $540 = 2^2cdot 3^3cdot 5$ and $72 = 2^3cdot 3^2$, so their GCD is $2^2cdot 3^2 = 36$ and their LCM is $2^3cdot 3^3cdot 5 = 1080$. The point is that if a number $d$ is to divide both $a$ and $b$, the power to which each prime appears in the factorization of $d$ cannot exceed the mininum power to which it appears in the factorization of both $a$ and $b$. A similar analysis works for the LCM.
$endgroup$
For example, $540 = 2^2cdot 3^3cdot 5$ and $72 = 2^3cdot 3^2$, so their GCD is $2^2cdot 3^2 = 36$ and their LCM is $2^3cdot 3^3cdot 5 = 1080$. The point is that if a number $d$ is to divide both $a$ and $b$, the power to which each prime appears in the factorization of $d$ cannot exceed the mininum power to which it appears in the factorization of both $a$ and $b$. A similar analysis works for the LCM.
answered Feb 17 at 14:32
rogerlrogerl
18k22748
18k22748
$begingroup$
Thank You so much!!! I got it! !
$endgroup$
– Jessi Jha
Feb 20 at 3:21
add a comment |
$begingroup$
Thank You so much!!! I got it! !
$endgroup$
– Jessi Jha
Feb 20 at 3:21
$begingroup$
Thank You so much!!! I got it! !
$endgroup$
– Jessi Jha
Feb 20 at 3:21
$begingroup$
Thank You so much!!! I got it! !
$endgroup$
– Jessi Jha
Feb 20 at 3:21
add a comment |
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