- #1

- 45

- 0

but what about sqrt(2 + sqrt(2)) or the iteration sqrt(2 + sqrt(2 + sqrt(2))).

the question is how to construct a line with length sqrt(sqrt(2)) i guess(beginning with lines of length 1), but i am not sure.

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- Thread starter raphael3d
- Start date

- #1

- 45

- 0

but what about sqrt(2 + sqrt(2)) or the iteration sqrt(2 + sqrt(2 + sqrt(2))).

the question is how to construct a line with length sqrt(sqrt(2)) i guess(beginning with lines of length 1), but i am not sure.

- #2

- 445

- 5

but what about sqrt(2 + sqrt(2)) or the iteration sqrt(2 + sqrt(2 + sqrt(2))).

the question is how to construct a line with length sqrt(sqrt(2)) i guess(beginning with lines of length 1), but i am not sure.

Just as you can construct a line of length root2 by drawing a triangle with sides 1 and 1, you can then extend the line of root2 to e.g. 5 + root2 by adding a segment of length 5 to the end, or convert it into root(root2 + 1) by drawing a perpendicular of length 1 on the end of the root2 line and creating a new hypotentuse.

- #3

- 45

- 0

- #4

- 445

- 5

Ah yes, my bad. I thought your way was right, but somehow managed to convince myself it was wrong :(

- #5

- 45

- 0

but you gave me some new ways to think about this...

maybe there are some other viewpoints?

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