Can any one please reply with the method of finding the l.c.m of $\frac{\pi}{2}$ and $\frac{\pi}{5}$. i arrived at the answer = $\frac{\pi}{10}$ ; but my book says that that the answer is $\pi$.
How to find lcm of $\frac{\pi}{5}$ and $\frac{\pi}{2}$ ? With paper pencil approach
4.8k Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtThere are 3 best solutions below
On
I take it that by a "multiple" of, say, $\pi/5$, we mean a number of the form $(\pi/5)r$, with $r$ an integer. Then by "least common multiple of $\pi/5$ and $\pi/2$" we must mean the smallest positive number $\alpha$ such that there are integers $r$ and $s$ with $(\pi/5)r=(\pi/2)s=\alpha$. This will be achieved by finding the smallest possible positive integers $r$ and $s$ such that $(\pi/5)r=(\pi/2)s$.
Now a little algebra turns that equation into $2r=5s$. Can you see that the smallest positive integers $r,s$ satisfying this equation are $r=5,s=2$? Well, those values give you $\alpha=\pi$, so that's the least common multiple you're trying to find.
On
For your question you need to understand the core concept of lcm
So what is lcm? Few of our teachers teach that while adding and subtracting a fraction we need to use lcm to equalize the denominator but that not correct
for example take 5/pi + 2/3 our teachers say that lcm of pi,3 is 3pi but that not correct, lcm of pi,3 does not exist
Lcm means the frequency of occurring the same multiples ex take 3,2 lcm of 3,2 is 6 that means 6 is the least multiple which can come in both the tables of 3,2 (Once check it )
while coming to the previous example lcm of $(3,\pi)$ does it have any common multiples It is impossible to have common multiples for a rational number and an irrational number
BTW Coming to your question $\pi/5 +\pi/2$ you need to be aware of a formula which is lcm of $a/b,c/d$ is lcm(a,c)/hcf(b/d) So by applying this rule lcm of $(\pi,\pi)$ is $\pi$ because $\pi$ is common multiple and hcf of $(5,2)$ is $1$ because they do not have any common factors so the answer is $\pi/1$ which is $\pi$
It is the same as finding $\DeclareMathOperator{\lcm}{lcm}\lcm\Bigl(\dfrac15,\dfrac12\Bigr)$. Now $$\lcm\Bigl(\frac1a,\frac1b\Bigr)=\frac1{\gcd(a,b)}.$$ Thus, one obtains $\;\lcm\Bigl(\dfrac\pi5,\dfrac\pi2\Bigr)=\pi\lcm\Bigl(\dfrac15,\dfrac12\Bigr)$.
As $\gcd(2,5)=1$, $\lcm\Bigl(\dfrac15,\dfrac12\Bigr)=1$, and indeed the smallest multiple of $\dfrac12$ which is equal to a multiple of $\dfrac15$ is: $$2\cdot\frac12=1=5\cdot\frac15,$$ and the lcm sought for is $\pi$.