Inequalities and logical statements: In a mathematical system, the following two statements relating to real number values for a, b, c, and d are (always) true: a < b and c < d .

Question

Then, Which of the following four statements are therefore always true:

I) (a + c) < (b + d) II) |a – c| < |b – d| III) ac < bd IV) (a – d) < (b – c)

Choices:

A) I, II, III, and IV B) I, II, and IV only C) I and III only D) I and IV only E) I only

Answer 1

Try plugging in numbers and test them out.

For example:

Let a = 0, b = 1, c = 2, and d = 3

Then:

Choice 1:

(a + c) < (b + d)

(0 + 2) < (1 + 3)

2 < 4 True

Choice 2:

|a - c| < |b - d|

|-2| < |-2|

2 < 2 False

Choice 3:

ac < bd

0 < 3 True

Choice 4:

(a - d) < (b - c)

-3 < -1 True

For now, only choice 2 is out.

Try making a = 2, b = 3, c = 0, and d = 1.

Choice 1:

(a + c) < (b + d)

2 < 4 True

Choice 3:

ac < bd

0 < 3 True

Choice 4:

(a - d) < (b - c)

2 < 3 True

This time, choose some negatives:

Let a = -2, b = 0, c = -1, and d = 3

Choice 1:

(a + c) < (b + d)

-3 < 3 True

Choice 3:

ac < bd

2 < 0 False

Choice 4:

(a - d) < (b - c)

-5 < 1 True

Now, we have found counter examples for both choices 2 and 3.

Lets look at choice 1 logically:

a and c are always smaller than b and d. The sum of 2 smaller numbers will always be less than the sum of two larger numbers. This choice will always be true.

Lets look at choice 4 logically:

A smaller number minus a bigger number is always going to be less than a bigger number minus a smaller number. Even if a > c and b > d, then since b > a, b is also greater than c and the difference will be positive. However, if a > c, then d can only either be equal to a or greater than a. So the difference can only by 0 or negative. So the statement will be true.

So choices 1 and 4 will always be true.

Answer 2

I is true, since we can add constants to inequalities. If a< b, we can add c to both sides to get a+c< b+c. And since c< d, we can add b to both sides to get b+c< b+d. So we have a+c< b+c< b+d.

II is asking whether the distance from a to c must be less than the distance from b to d. But we can take b, c, d to be clustered close together, and a very far from them, such as c=0, b=1, d=2, a=-100. So this needn't be true. B is eliminated.

For III, let's examine a simple case, where c=0. This means d is positive, and we're asking whether 0<bd. Since we can take b to be negative, this needn't be true either.

And for IV, we can rearrange this inequality into a+c< b+d. This is the inequality from I, which we already proved true. So only I and IV are true, and the answer is D.

Answer 3

Strategy: These are logical-conditional statements. False-logically means any statement that can ever be false, whereas true-logically means any statement that is always true. Thus, seek conditions which may falsify the test statement. Often, these are “extremes” of one kind or another. It does no good to find conditions which satisfy the statement – these do not establish logical-trueness for the statement!

Tactics: Only addition works for combining comparably 'ordered' inequalities, such as obtain here. Thus I and IV (which is an equivalent statement to I) are true. Both statements II and III are violated by the following number-line assignments (draw a number line to observe): A = -4, C = 0, B = 3, D = 4 .

(The || absolute value symbol doesn't help to make that statement any truer!)

Essentially, the addition operation (on inequalities) preserves the direction and magnitude of offsets of the two sides of the expression from the origin. The subtraction operation (on inequalities) compares the magnitude of the separation each inequality expresses across the “<” sign; when the subtracted equation is (= may be!) further from being an equality, the final expression is false (as above).

The multiplication and division operations (on inequalities) may flip values between negative or positive or zero, depending on the signs of the individual variables as well, which is NOT controlled in the specified initial statements, so these operations falsify the resultant inequality (remember, “may falsify” == falsify for logic-conditional statements!).