0+0+0... = 1

[Anonymous No.10385233]

[Anonymous No.10385566]

>>10385233

[Anonymous No.10385581]

[eqn]0\,+\,0\,+\,0\,+\,\cdots \;=\; \lim_{n\;\to\;\infty}\, \sum_{i\;=\;1}^n \frac1n \;=\; \lim_{n\;\to\;\infty}\,\frac nn \;=\; \lim_{n\;\to\;\infty}\, 1 \;=\;1[/eqn]

[Anonymous No.10385653]

>>10385233
It depends on the characteristic of your field. If it is 2, then:
[eqn]1 + 1 = 0[/eqn]
[eqn]1 + 1 = 0 \tag{+}[/eqn]
[eqn]\overline{1 + 1 + 1 +1 = 0 + 0}[/eqn]
[eqn]\frac{1+1+1+1}{4}=\frac{0+0}{4}[/eqn]
[eqn]\frac{1+1+1+1}{1+1+1+1}=\frac{0}{4}+\frac{0}{4}[/eqn]
[eqn]1=0+0[/eqn]

If the characteristic of your field is n you can generalize:
[eqn]0_1+0_2+0_3+\dots+0_n=1[/eqn]

But if there isn't any n such that adding [math]0[/math] n times, you must say that the characteristic of the field is 0.