Determine if the biconditional statements are true or false.
1. Any whole number is divisible by 3 if and only if the sum of its digits is divisible by 3.
2. 1/x > 0 if and only if x ≥ 0.
3. 2 is a prime if and only if 7 is even.
1. yes --- https://www.apronus.com/math/threediv.htm
2. no try x = 0.5
3. makes no sense to me
in number 2, if you try it with 0.5 then 2 > 0 and 0.5 ≥ 0 then it is true? while if i try it on x = 0, 1/0 is undefined, and 0 ≥ 0 is true, therefore the biconditional statement is false?
To determine if the biconditional statements are true or false, we need to analyze both the forward and backward implications.
1. Any whole number is divisible by 3 if and only if the sum of its digits is divisible by 3.
To check the forward implication (if a whole number is divisible by 3, then the sum of its digits is divisible by 3), we can examine different examples. Consider the following number: 135. The sum of its digits (1+3+5) equals 9, which is divisible by 3. If we divide 135 by 3, we get a quotient without any remainder. This supports the forward implication.
To check the backward implication (if the sum of the digits of a whole number is divisible by 3, then the whole number itself is divisible by 3), we can also analyze examples. For instance, take the number 72. The sum of its digits (7+2) equals 9, which is divisible by 3. If we divide 72 by 3, we again obtain a whole number without any remainder. This supports the backward implication.
Since both the forward and backward implications hold true, the biconditional statement is true. Therefore, any whole number is divisible by 3 if and only if the sum of its digits is divisible by 3.
2. 1/x > 0 if and only if x ≥ 0.
For the forward implication (if 1/x > 0, then x ≥ 0), we need to examine the inequality itself. If 1/x is greater than 0, it means the reciprocal of x is positive. For any positive number, the value of x must be greater than 0. Thus, the forward implication holds true.
Now, let's analyze the backward implication (if x ≥ 0, then 1/x > 0). If x is greater than or equal to 0, then the reciprocal of x will still be positive or 0. Since 0 is neither positive nor negative, we exclude it from being considered as a counter-example. Thus, the backward implication also holds true.
Consequently, the biconditional statement is true. Therefore, 1/x > 0 if and only if x ≥ 0.
3. 2 is a prime if and only if 7 is even.
To verify the truth of the biconditional statement, we will examine both implications individually.
For the forward implication (if 2 is a prime, then 7 is even), we know that a prime number is defined as a number greater than 1 with only two distinct positive divisors: itself and 1. However, 7 does not satisfy the criterion for being divisible solely by 1 and itself, as it can be divided by 1 and 7. Consequently, 2 being a prime does not imply that 7 is even.
Next, we check the backward implication (if 7 is even, then 2 is a prime). Even numbers are multiples of 2, but 7 is an odd number, as it cannot be divided evenly by 2. Therefore, 7 being even does not imply that 2 is a prime.
Since neither the forward nor backward implication holds true, the biconditional statement is false. Therefore, 2 is not a prime if and only if 7 is even.