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Am 26.08.2024 um 02:15 schrieb Moebius:We may do this (from a logical point of view) JUST BECAUSE we assumed (at the start of the proof) the existence of at least one such unit fraction.Am 25.08.2024 um 23:18 schrieb Jim Burns:On 8/25/2024 3:35 PM, WM wrote:Btw. To show this we [usually] don't refer to "the smallest unit fraction">Each non.empty set of ordinals holds a first.>
Like each set of unit fractions.
Fascinating.
>
On the other hand, if u is a unit fraction, then 1/(1/s + 1) is a smaller one.
>For each unit.fraction u>
the claim that u is smallest is proved false
by counter.example ⅟(1+⅟u)
Indeed! :-)
WM (<- a constant denoting "the smallest unit fraction").
I mean, we don't introduce (define) WM ("the smallest unit fraction") by a "definition" just to show (afterwards) that it does not exist: ~Ex e SB: s = WM.
Of course, in a proof by contradiction we may proceed the following way:
Assume that there IS _a_ smallest unit fractions. i.e.
Es e SB: As' e SB\{s}: s < s' .
Let wm (<- an arbitrary name or parameter or ...) be such a unit fraction.
With other words:
wm e SB & As' e SB\{wm}: wm < s'
and hence
As' e SB\{wm}: wm < s'.
Now from wm e SB we get that 1/(1/wm + 1) e SB and 1/(1/wm + 1) < wm, contradicting
As' e SB\{wm}: wm < s' .
qed.
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