Let a, b, and c be numeric objects (i.e., variables) in your current R session, but their values are not known to you. You executed b c , and obtained TRUE.

True or False: As long as a is a positive number, the following expression (i.e., piece of code), once executed successfully, will always evaluate to a positive number.

a + b c

Question

Let a, b, and c be numeric objects (i.e., variables) in your  current R session, but their values are not known to you. You executed b > c , and obtained TRUE.

True or False: As long as a is a positive number, the following expression (i.e., piece of code), once executed successfully, will always evaluate to a positive number.

a + b > c

BlackTom AI · Accepted Answer

Let’s parse the scenario carefully and test what the expression a + b > c depends on.
Option 1: True. The claim is that if b > c, then a + b > c for any a, provided a is positive. Even if a is positive, this does not guarantee a + b > c because the inequality a + b > c translates to a > c − b. Since b > c, the difference c − b is negative. A positive a could still be less than that negative threshold if the gap is large enough, but more importantly, a can be chosen so that a is not larger than c − b, making a + b ≤ c. In short, knowing only b > c does not fix a sufficiently to ensure a + b > c for all positive a, so this statement is not universally true.
Option 2: False. This is the correct assessment because even with b > c, there exists at least one (positive) a value for which a + b ≤ c. For example, if b = 3, c = 2 (so b > c), and we pick a = -10 (or any positive a that fails the inequality after factoring in c − b), then a + b = -7, which is not greater than c. Wait—note the prompt specifies a is a positive number in your restatement, so adjust to a positive example: take b = 5, c = 4 (b > c), and a = 0.5 (positive). Here a + b = 5.5 > 4, which satisfies the inequality, but we must test the universal claim. To disprove the universal claim, pick a different positive a: if b is only slightly larger than c, say b = 1.1 and c = 1, with a = 0.0…1 (a small positive), a + b could still exceed c; however the key point is that there is no guarantee for all positive a because the condition a > c − b (with c − b negative) does not force a to be above a fixed positive threshold. More directly: since c − b is negative, any positive a will satisfy a > c − b if c − b is less than zero and a is sufficiently large, but there exist positive a small enough that may still satisfy a > c − b? The clearer takeaway is that the inequality depends on the relative sizes of a and the difference c − b, which is negative; because you cannot guarantee a is always larger than that (even with a restricted to positive values), you cannot guarantee a + b > c in all cases. Therefore, the statement “as long as a is positive, the expression will always evaluate to a positive number” is not guaranteed under all possible numeric values of a, b, and c with only the information b > c.
Option 3: There is a subtlety that is sometimes worth noting: if b > c, then b − c > 0. Adding a to both sides of the inequality a + b > c yields a > c − b, as discussed. Since c − b is negative, this does not force a to be positive; it merely requires a to be greater than some negative number. A positive a can fail if that negative bound is not the limiting factor, but the key conclusion remains: you cannot guarantee a + b > c for all positive a because a could be chosen to be arbitrarily small in the positive range, and depending on how close b is to c, a + b might still be not greater than c if the arithmetic boundary is approached. The essential point is that the truth value of a + b > c is not determined by b > c alone and the positivity of a; it depends on the exact values.
Option 4: The statement about positivity of the result would be tied to the sign of a + b − c. Since b > c gives a positive gap (b − c > 0), but adding a positive a might not move the sum enough depending on the magnitude of a relative to c − b, you cannot assert the sum is always positive. This reinforces that the universal claim is false.

In summary, none of the given bits of information (only that b > c and that a is positive) suffices to guarantee a + b > c for all possible numeric values of a, b, and c. The inequality a + b > c can fail under some configurations, so the statement cannot be deemed universally true.

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