Why is 1/(|x-4|) the same as |x-4|>1?
(1 votes)
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You misrepresent the equivalence relation from the neckline.
This applies as you can easily multiply on both sides with carry(x-4)
It’s not the same thing. For x=4, the left term is indeterminate and the right term has the truth value “false”.
Otherwise you can multiply with |x – 4|, which is always positive.
from 1 / |x – 4| < 1 is then 1 < |x - 4|
Doesn’t the truthful for the left unequalization also exist as “false” if the term 1/|x-4| is indefinite, so if 0 is in the denominator? Thus would follow from false wrong, which would make the equivalence relationship “true”.
I am hard to assign a truth value to the statement 1/0 < 1.
In the logic of today’s floating point processors, however, 1/0 < 1, 1/0 = 1 and 1/0 > 1 are all three false statements.
That’s what I’d interpret. If you start by True as defining, a statement is wrong if it is not true. These statements are true only for certain clearly defined values. Since 1/0 does not describe one of these clearly defined values, the above-mentioned statements are considered to be incorrect. Whether the term 1/0 itself can be attributed a unique value at all. As far as my interpretation of the matter.
Take a look
1 < 1* |x-4|
the left side should be smaller than the right
you can turn it around
you right is bigger than the left, what to
|x-4| > 1
guide
That’s the EQ!