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Talk:Conservapedia insights (view source)
Revision as of 19:55, July 27, 2021
, 19:55, July 27, 2021→Proof that π does not contain π:
If we can agree on the above, then the following should be straight forward:
If we can agree on the above, then the following should be straight forward:
#If π starts again at <i>n</i>, then x<sub>0</sub> = x<sub>n</sub> = 3, x<sub>1</sub> = x<sub>n+1</sub> = 1, x<sub>2</sub> = x<sub>n+2</sub> = 4, x<sub>3</sub> = x<sub>n+3</sub> = 1 and so on. "And so on" means that x<sub>k</sub> = x<sub>k+n</sub> for each natural number <i>k</i>, i.e.,. <i>k ∈ <b>N</b><sub>0</sub></i>. Andy, you seemed to have no qualms with this step.
#If π starts again at <i>n</i>, then x<sub>0</sub> = x<sub>n</sub> = 3, x<sub>1</sub> = x<sub>n+1</sub> = 1, x<sub>2</sub> = x<sub>n+2</sub> = 4, x<sub>3</sub> = x<sub>n+3</sub> = 1 and so on. "And so on" means that x<sub>k</sub> = x<sub>k+n</sub> for each natural number <i>k</i>, i.e.,. <i>k ∈ <b>N</b><sub>0</sub></i>. Andy, you seemed to have no qualms with this step.
#What happens if <i>k</i> equals <i>n</i>? What is the value of x<sub>n + n</sub>? According to the step above, x<sub>n + n</sub> =x<sub>n</sub>. But x<sub>n</sub>=x<sub>0</sub>=3 ! So, x<sub>n + n</sub> = 3. That's true for all multiples of <i>n</i>: if <i>w ∈ <b>N</b><sub>0</sub></i>, then x<sub>w n</sub> = x<sub>(w-1) n</sub> = .... = x<sub>2 n</sub> = x<sub>n</sub> = x<sub>0</sub> = 3.
#What happens if <i>k</i> equals <i>n</i>? What is the value of x<sub>n + n</sub>? According to the step above, x<sub>n + n</sub> =x<sub>n</sub>. But x<sub>n</sub>=x<sub>0</sub>=3 ! So, x<sub>n + n</sub> = 3. That's true for all multiples of <i>n</i>: if <i>w ∈ <b>N</b><sub>0</sub></i>, then x<sub>w n</sub> = x<sub>(w-1) n</sub> = .... = x<sub>2 n</sub> = x<sub>n</sub> = x<sub>0</sub> = 3.
#What about x<sub>w n +k</sub> (<i> 0 ≤ k < n </i>)? Again, x<sub>w n +k</sub> = x<sub>(w-1) n +k</sub> = x<sub>(w-2) n +k</sub> = x<sub>2 n +k</sub> x<sub>n +k</sub> = x<sub>k</sub>
#What about x<sub>w n +k</sub> (<i> 0 ≤ k < n </i>)? Again, x<sub>w n +k</sub> = x<sub>(w-1) n +k</sub> = x<sub>(w-2) n +k</sub> = x<sub>2 n +k</sub> x<sub>n +k</sub> = x<sub>k</sub>