L.C.M. of two irrational numbers always exists ?

Number Theory Level 2

$\huge \color{#D61F06}{\pi \times e}, \color{#20A900}{\pi}$

Find the lowest common multiple (L.C.M.) of the given above two numbers ?

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L.C.M. does not exist.

e

\pi

None of the above

\pi \times e

1 solution

Abhishek Sharma
May 8, 2015

If LCM of $a$ and $b$ is $L$ , then $L=a\times n=b\times m$ where $n$ and $m$ are integers.

Now if LCM of $\pi \times e$ and $\pi$ exists then $e=\frac{m}{n}$ . But this can't be true as $e$ is an irrational number and $\frac{m}{n}$ is rational. Our assumption was wrong therefore LCM of $\pi \times e$ and $\pi$ does not exist.

Why can't the following be true...

LCM×HCF=PRODUCT OF THE TWO.,

But HCF of π&eπ is π =>LCM=product÷ hcf

So LCM =( π×π×e) ÷ π = πxe

Ashwin Kumar - 6 years ago

You are wrong about HCF (logic similar to solution).

Abhishek Sharma - 5 years ago

Generally in maths when you come across an obstacle you define something - for example - complex numbers to solve quadratic equations.

Then, why does it not make sense for a generalisation of the LCM to be defined where LCM is a real multiple (not integer multiple) of the two numbers

Star Light - 5 years ago

L.C.M. of two irrational numbers always exists ?

Join the Brilliant Classes and enjoy the excellence.

1 solution

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