Mathematics
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OpenStudy (unklerhaukus):
\[ x \divideontimes x =1\]
\[x=?\]
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OpenStudy (anonymous):
what is that sign? :S
OpenStudy (unklerhaukus):
divide on times
OpenStudy (unklerhaukus):
a bit like ±
OpenStudy (anonymous):
oh! thanks..
OpenStudy (unklerhaukus):
so there can only be two solutions right?
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OpenStudy (unklerhaukus):
come on someone , take a guess,,,
OpenStudy (diyadiya):
\[x \div x=1\]
=D
OpenStudy (diyadiya):
\[x \times x =x^2\]
this is all i know
OpenStudy (unklerhaukus):
\[1 \divideontimes 1 =1\] True
because \[1 \div 1 =1, \qquad and \qquad 1 \times 1=1\]
OpenStudy (unklerhaukus):
what is the other solution?
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OpenStudy (anonymous):
I think there is only one solution.
OpenStudy (anonymous):
0?
OpenStudy (anonymous):
No, can't be....
OpenStudy (diyadiya):
No 0/0 = indeterminate
OpenStudy (unklerhaukus):
\[\frac{0}{0} ≠ 0\]
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OpenStudy (anonymous):
Yes, I never thought of that
OpenStudy (anonymous):
I agree with jasonchutko, only one solution.
OpenStudy (unklerhaukus):
what about \[-1\]
OpenStudy (diyadiya):
Oh cool!
OpenStudy (anonymous):
-1 * -1 = 1
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OpenStudy (anonymous):
Very true...
OpenStudy (diyadiya):
i've never seen that sign before
OpenStudy (anonymous):
Me too..
OpenStudy (anonymous):
So the answer is one?
OpenStudy (diyadiya):
1 and -1
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OpenStudy (unklerhaukus):
\[x=±1\]
OpenStudy (anonymous):
But -1 * -1 = 1!!
OpenStudy (unklerhaukus):
which satisfies the original equation
\[x \divideontimes x =1\]
OpenStudy (anonymous):
Oh, yes I thought it had to be equal to the same number ... my mistake.
OpenStudy (unklerhaukus):
there is only one solution to \[x \divideontimes x =x\]\[x=1\]