Prove that the Greatest Integer Function f: R -> R given by f(x) = [x], is neither one-once nor onto, where [x] denotes the g
reatest integer less than or equal to x.
1 answer:
F: R → R is given by, f(x) = [x]
It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1
So, f(1.2) = f(1.9), but 1.2 ≠ 1.9
f is not one-one
Now, consider 0.7 ε R
It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ε R such that f(x) = 0.7
So, f is not onto
Hence, the greatest integer function is neither one-one nor onto.
The answer was quite complicated but I hope it will help you.
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6, 18
Step-by-step explanation:
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Let
x--------> the number of gray bricks
y--------> the number of red bricks
we know that
------> inequality 1
<em>-----> </em>inequality 2
therefore
<u>the answer is</u>
x > 0
0.45x+0.58y≤200
<span>−8x+5−2x−4+5x
= -5x + 1
so when x = 2 then
</span>-5x + 1
= -5(2) + 1
= -9
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