- #71
Mute
Homework Helper
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Hurkyl said:Let a be a number such that a^a=a/a
a^a=a/a
a^(a-1)=1/a
a^(a-1)=a^-1
a-1=-1
a=0.
Ah, the classic fallacies of forgetting your hypotheses, and reversing the flow of logic.
The very premise of your problem requires a to be a positive number. (0 is not a positive number)
Also, what you have proved is
If a is a (positive) number satisfying [itex]a^a=a/a[/itex], then a = 0which is very different from
If a=0, then [itex]a^a = a/a[/itex].
Nevermind the fact that he/she divided by a in the derivation, and took a logarithm of both sides (which also happened to remove the a = 1 solution).
dimension10 said:0^5=1*0*0*0*0*0
0^4=1*0*0*0*0
0^3=1*0*0*0
0^2=1*0*0
0^1=1*0
0^0=1
Thus,
0^0=1.
0^5=17*0*0*0*0*0
0^4=17*0*0*0*0
0^3=17*0*0*0
0^2=17*0*0
0^1=17*0
0^0=17
Thus,
0^0=17?