When the ceiling gets a fracture
True or false :
The value of the expression ( ⌈ { x } ⌉ ) x is either 0 or 1 for all real values of x .
Notations :
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⌈ ⋅ ⌉ denotes the ceiling function .
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{ ⋅ } denotes the fractional part function .
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2 solutions
Isn't 0^0 is equal to 1 ?
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No, 0 0 is indeterminate. Check out the debate and the links to these discussions here
What? 0 0 is indetermibate as far as I know. Refer here
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I used calculator and then google at the time of solving this question as i was not sure about the value but both reveled same result that is 1. So i was totally confused.
It is true for all real values x, cause at x=0, it is undefined, so it 1 and 0 at the same time, thus, the statement is true.
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For now, 0 0 remains indeterminate, not undefined. Nobody really knows what exactly is 0 0 , but the debates usually say that it is either 1 or 0 . Although mathematicians assign it a value of 1 , it is merely just an assumption for convenience in proving formulas. For all we know, we might have been wrong about it all this while.
If it is indeterminate, we cannot say for sure what the value is, because we don't know and cannot be certain about the value. Therefore, we cannot say that this statement is true
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Clearly the expression is either 0 or 1 (but not both) for all real values x (except x=0). At x=0 the expression is indeterminate, as Ashish has shown. But, at x=0 we can say that the expression becomes 0 and 1 at the same time, thus, it is indeterminate as a number cannot be two different numbers, therefore, the statement is still true for all real values. The "or" is not exclusive, unless stated otherwise. In some languages, there exist a word for the "exclusive or" and another word for the "inclusive or".
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@Mateo Matijasevick – You're not getting the idea. How are you so sure that 0 0 equals to 0 or 1 ? It's indeterminate because we cannot show that it is 0 , 1 , some other value or undefined
@Mateo Matijasevick – Again why are you concludong that 0 0 is 1 or 0 ?
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@Ashish Menon – I'm concluding that it is 1 AND 0 at the same time, therefore, it is indeterminate.
If the expression becomes 0 and 1 at the same time, it is indeterminate. Therefore, the statement is true.
I got it thanks @Hung Woei Neoh :)
Let me ask you something: if an expression becomes two values at the same time, can we say that is indeterminate?
Undefined is 1 or 0? Plz elaborate. Undefined is not always 1 or 0. Again this reason is stil invalid because the question states that it is either 1 or 2, it cant be both at the same time.
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The "or" in logic is defined as the disjunction, it is not exclusive, so it can be both at the same time. Undefined doesn't mean always 1 or 0, but if x is 1 and 0, then it is undefined.
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@Mateo Matijasevick – I believe that the question is correct though. Plz explain in a bit detail.
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@Ashish Menon – Clearly the expression is either 0 or 1 (but not both) for all real values x (except x=0). At x=0 the expression is undefined, as you have shown. But, at x=0 we can say that the expression becomes 0 and 1 at the same time, thus, it is undefined as a number cannot be two different numbers, therefore, the statement is still true for all real values. The "or" is not exclusive, unless stated otherwise. In some languages, there exist a word for the "exclusive or" and another word for the "inclusive or".
Counterexample: x = − 1
You end up with 0 − 1 which is undefined.
The exprrssion is not value for x = 0 .
At x = 0 , ⌈ { x } ⌉ = 0 . So, ( ⌈ { x } ⌉ ) x = 0 0 = undefined.
Since this expression does not stand true for x = 0 , the answer is False .