Fundamental period of two functions I have the function
$$f(t)=\sin(t)+\sin(\sqrt2t)$$
I would like to calculate the fundamental period of $f(t)$. I know that the period of $\sin(t)$ is $2\pi$ and the period of $\sin(\sqrt2t)$ is $\sqrt2\pi$. I sense that I must work out the lcm of $2$ and and $\sqrt2$ but I'm unsure on how to do this. 
 A: Your equation boils down to solving for some $a,b\in\Bbb{Z}$
$$2a\pi=\sqrt{2}b\pi$$
But this is clearly not possible because then $b=\sqrt{2}a$,but $\sqrt{2}$ is irrational and $b$ isn't.So $f(t)$ is not periodic.
A: When two numbers have an lcm ‒ say $18$ and $24$ ‒ there are integers by which you can multiply both of them to get the lcm, thus:
\begin{align}
18 \times 4 & = 72 \\
24 \times 3 & = 72 \\[10pt]
\text{so } \operatorname{lcm}(18,24) & = 72
\end{align}
and those two multipliers are $4$ and $3$. What would be the multipliers in the case of $2$ and $\sqrt 2\text{ ?}$
\begin{align}
2\times n & = \operatorname{lcm} \\
\sqrt 2 \times m & = \operatorname{lcm}
\end{align}
so $\dfrac m n = \dfrac 2 {\sqrt 2} = \sqrt 2$ and $m$ and $n$ are integers and the fraction $m/n$ is in lowest terms.
A bit of algebra shows that $\dfrac{2n-m}{m-n} = \sqrt 2$ and then this expresses that fraction ins lower terms. That is impossible if it was already in lowest terms.
Conclusion: there can be no such integers; therefore $\sqrt 2$ is an irrational number.
Further conclusion: There is no such lcm.
And the punch line: Therefore, this function is not periodic.
