roastedchips wrote:
We define that [x] is the least integer greater than of equal to x. [x+1]=?
1) [x] =1
2) [2x]=1
Let us analyze each statement
Statement1:
\([x]=1\)
\(0<x<=1\)
\(1<x+1<=2\) (Adding 1 throughout the inequality)
We know that x+1 will be greater than 1 and less than/equal to 2. So the least integer greater than or equal to\(x+1\) is 2.
Hence sufficient.
Statement 2:
\([2x]=1\)\(0<2x<=1\)
\(0<x<=0.5\) (Dividing by 2 throughout the inequality)
\(1<x+1<=1.5\) (Adding 1 throughout the inequality)
We know\(x+1\) will be greater
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