Describe the binary expression of $4^n$

The answer is: a 1 followed by n zeroes. Why?

I answered $110^n$

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I'm reading my commutative algebra notes and there is a remark that if
$f: M \to M^\prime$ and $g: N \to N^\prime$ are $R$-module homomorphisms then
$$ f \otimes g: M \otimes N \to M^\prime \otimes N^\prime$$
$$ m \otimes n \mapsto f(m) \otimes g(n)$$

is an $R$-module homomorphism. The remark also notes that this map can and should be constructed rigorously.

By the universal property of the tensor product $(T=M \otimes N, b)$ we have that for the bilinear map
$$h: M \times N \to M^\prime \otimes N^\prime$$
$$ (m,n) \mapsto f(m) \otimes g(n)$$

there exists a unique $R$-module homomorphism $l : M \otimes N \to M^\prime \otimes N^\prime$ such that $l \circ b = h$.

We have $h((m,n)) = f(m) \otimes g(n) = l(b(m,n)) = l(m \otimes n)$. Hence we have that $f \otimes g := l$ is an $R$-module homomorphism $m \otimes n \mapsto f(m) \otimes g(n)$.

all in all you're just a-
-nother brick in the wall

@book{18647,
title = "General irreducible Markov chains and non-negative operators.",
address = "Cambridge",
publisher = "Cambridge Univ.Press",
year = "1984"
}