Morgoth wrote:is x < 1 ?

1) x^-1/3

2) x^-2

OA after some posts.

Where are the statements?

In logic a statement is a declarative sentence that is either true or false. Strawson however advocated the use of the term statement (in preference to proposition) and for it to be such that two declarative sentences make the same statement if they say the same of the same thing. Thus the term "statement" may to refer to a sentence or something made (expressed) by a sentence. In either case they are purported truth bearers.

Examples of sentences that are (or make) statements:

"Socrates is a man."
"A triangle has three sides."
"Paris is the capital of Japan."
The first two (make statements that) are true, the third is (or makes a statement that is) false.

Examples of sentences that are not (or do not make) statements:

"Who are you?"
"Run!"
"Greeness perambulates"
"I had one grunch but the eggplant over there."
The first two examples are not declarative sentences and are therefore (or do not make) statements. The third and forth are declarative sentences but, lacking meaning, are neither true nor false and therefore are not (or do not make) statements.

In some treatments the term "statement" is introduced in order to distinguish a sentence from its information content. A statement is regarded as the information content of an (information-bearing) sentence. Thus, a sentence is related to the statement it bears like a numeral to the number it refers to. Statements are abstract, logical entities, while sentences are grammatical ones.

Good to see u back after some time. Welcome back Morgoth!

Regards,
Cramya

Morgoth wrote:is x < 1 ?

1) x^-1/3 < 1

2) x^-2 < 1

1) x^-1/3 < 1:

Consider the domains of x < 0, 0 < x < 1, and x >= 1

Where x < 0, x^1/3 is -ve, so its reciprocal is also negative, and therefore less than one.

For 0 < x < 1, 0 < x^1/3 < 1, so x^-1/3 is > 1
For x > 1, x^1/3 > 1 so x^-1/3 is < 1
When x = 0 or x = 1, x^-1/3 = 1

So if x^-1/3 < 1, either x > 1 or x < 0, INSUFFICIENT

2) x^-2 < 1

If x > 1, x^2 > 1, so x^-2 < 1
If 0 > x > 1, then 0 < x^2 < 1, so x^-2 > 1
If -1 < x < 0, then 0 < x^2 < 1, so x^-2 > 1
If x < -1 then x^2 > 1, so x^-2 < 1
If x = -1, x = 0, or x = 1 then x^2 = 1, so X^-2 = 1

So if x^-2 < 1, then x < -1 or x > 1, INSUFFICIENT

1 & 2: Since x = -2 or x = 2 would both satisfy 1 & 2 but not determine the question, it's E.

from stmt 1, x can be greater or smaller than 1
from stmt 2 x is less than -1 or greater than 1

using both 1 & 2 no unique solution
hence E