# The Notion of LCM

$$\bullet$$ LCM of any two non-zero rational numbers always exists.

$$\bullet$$ LCM of any non-zero rational and any irrational number never exists.

$$\bullet$$ LCM of any two irrational numbers may or may not exist.

$$\bullet$$ Also you can start out the discussion here on this note through comments.

Till then you can try to solve the set of such problems : IIT Foundation Classes

Note by Sandeep Bhardwaj
3 years, 2 months ago

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What is so cool about LCMs?

Staff - 3 years, 2 months ago

Can you post questions on Polynomials ? and Geometry ?

- 3 years, 2 months ago

What is the most accepted definition of LCM sir, which is in accordance with all integers, rationals and irrationals ? (As you may have known, most people like me are having a problem with the definition ).

Thanks in advance :).

- 3 years, 2 months ago

What are the conditions for two irrational numbers to have an LCM?

- 3 years, 2 months ago

According to my understanding :

LCM of two like irrational numbers always exists.

Now what I mean to say like here is :

Let any irrational number being $$\lambda$$. then another irrational number $$\alpha$$ will be like to $$\lambda$$ if $$\alpha=A \times \lambda$$ where $$A$$ is any non-zero rational number.

- 3 years, 2 months ago

But Sir i thought in your previous questions there are no LCM for two irrational numbers nor between irrational number and rational number.

- 3 years, 2 months ago

- 3 years, 2 months ago

Yeah, LCM exists in both the cases. in one case LCM is $$6\pi$$ and in the other case LCM is $$2e$$.

- 3 years, 2 months ago

Thank you very much sir, I was scared that my question was wrong! I love you sir! :P $$\huge \ddot \smile$$

- 3 years, 2 months ago

Thank you my dear, $$\huge \ddot \smile$$

- 3 years, 2 months ago

BTW, How were the questions sir?

- 3 years, 2 months ago

Sir U gave away the answer

- 3 years, 2 months ago

Yeah, I would Like to Know the same. :)

- 3 years, 2 months ago

Good explanation about LCM'S keep it up::)

- 3 years, 2 months ago