Relations proof

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[inland anchor]

well notice if you use induction, and assume it holds for $n-1$, that the product there is equal to $(R_{n-1} - 2)\cdot R_{n-1}$

[hasty mauve]

Pyrokinetics

[inland anchor]

then you can sub in

[wide obsidian]

Wht

[limpid sundial]

are you familiar with induction?

[wide obsidian]

we barely touched on it

but im aware we need to prove it for a base case and prove for n + 1

[limpid sundial]

that is indeed induction

what are your thoughts about n=0?

an "empty" product can be set to equal to 1

so n=0 also holds

[wide obsidian]

ive never seen that bridge looking notation but i assume its similar to sigma but with products

if n = 0

then it will go up to -1 times?

[limpid sundial]

$$ \prod _{k=1}^n x_k = x_1\cdot x_2\cdot \ldots\cdot x_n$$

[hasty mauve]

aL

[inland anchor]

$\prod_{n=a}^{b} f(n)$ is called an empty product if $a > b$ because there are no numbers that are at least $a$ and at most $b$
so this is often set to 1 just like an empty sum (the sum of no numbers) is usually set to 0

[hasty mauve]

Pyrokinetics

[wide obsidian]

#OK#

when n = 0 is there a concise way to show that the empty product will = 1

[inland anchor]

no, its by convention

[wide obsidian]

we never even seen the sum of product sign so i cant assume that the empty product is 1 without any proof

[limpid sundial]

in how many ways can you order a queue with no people in it?

1

don't worry about it too much, empty product = 1, empty sum = 0, figure it out when you are a phd student and are required to chase rabbits down rabbit holes

[wide obsidian]

Ok