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OpenStudy (anonymous):
Are you serious?
OpenStudy (skullpatrol):
Yes.
OpenStudy (anonymous):
Always True
OpenStudy (anonymous):
c=1/1c
OpenStudy (anonymous):
Not a Quadratic
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OpenStudy (isaiah.feynman):
Could use limits.. perhaps.
OpenStudy (isaiah.feynman):
@skullpatrol
OpenStudy (anonymous):
Multiply both sides by \(1\):\[
1c = c
\]Is this sufficient?
OpenStudy (anonymous):
You seem to be assuming the statement is true in order to prove it is true.
OpenStudy (anonymous):
Like this:\[
\lim_{x\to 1}x = 1=-1
\]
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OpenStudy (the_fizicx99):
Wouldn't that just be c = c ?
OpenStudy (skullpatrol):
$$\frac{a}{1}=a*\frac{1}{1}$$ because $$\frac{a}{b}=a*\frac{1}{b}$$.
$$1*\frac{1}{1}=1$$ because $$a*\frac{1}{a}=1$$.
$$1*1=1$$ because $$a*1=a$$.
Comparing the two equations above:
$$\frac{1}{1}=1$$.
$$\therefore \frac{a}{1}=a*\frac{1}{1}=a*1=a$$.
OpenStudy (the_fizicx99):
Exactly
OpenStudy (anonymous):
simple absract algebra !
just define the operation "multiple"
and define the identity "1"
OpenStudy (turingtest):
Indeed, multiplicative identity could be made to be zero in the right vector space, for instance.