1). Let G be the statement: " for all real numbers a and b, if a -|_a_| < 1/2 and b - |_b_| < 1/2 then |_a + b_| = |_a_| + |_b_|"
(the symbol is for the FLOOR of a and b)
A. is G true? prove it
b. state wether the CONVERSE, CONTRAPOSITIVE AND NEGATION ARE TRUE and prove.
Thanks for all your help.
(a-[a]) + (b-[b]) < 1/2 + 1/2 = 1
so, the fractional part of the sum will not cause it to exceed the next integer.
will think on the other parts, but maybe you have some ideas you can post in the meantime.
To determine whether the statement G is true, we need to prove it. Let's break it down into two cases:
Case 1: a and b are both integers.
Let's assume a = m and b = n, where m and n are integers. Now we need to prove |_m + n_| = |_m_| + |_n_|.
|m + n| represents the absolute value of the sum m + n. Since m and n are integers, their sum m + n will also be an integer. Therefore, |_m + n_| = |m + n| = m + n, as the absolute value of the sum is equal to the sum itself when dealing with integers.
Now consider the right-hand side, |_m_| + |_n_|. Here, |_m_| is the floor value of m, which is m since m is an integer. Similarly, |_n_| is n. Therefore, |_m_| + |_n_| = m + n.
We can see that both sides of the equation are equal: m + n = m + n. Hence, for integers a and b, the statement is true.
Case 2: a and b are not both integers.
Let's consider a = p + r and b = q + s, where p, q are integers and 0 <= r, s < 1.
Now we need to prove |_p + q + r + s_| = |_p + r_| + |_q + s_|.
|_p + q + r + s_| represents the floor value of p + q + r + s. We know that p + q + r + s will be an integer since p, q are integers and r, s are fractional parts.
Now let's consider the right-hand side, |_p + r_| + |_q + s_|. Since p and q are integers, |_p + r_| = |_p + r| = p + r, and |_q + s_| = |_q + s| = q + s, as the floor value of a real number which is not an integer is the integer part of that number.
So, |_p + r_| + |_q + s_| = p + r + q + s.
Therefore, |_p + q + r + s_| = |_p + r_| + |_q + s_|.
We can see that both sides of the equation are equal: |_p + q + r + s_| = p + r + q + s.
Hence, for real numbers a and b, the statement is also true.
In conclusion, we have shown that for all real numbers a and b, if a - |_a_| < 1/2 and b - |_b_| < 1/2, then |_a + b_| = |_a_| + |_b_|.
Now let's move on to the converse, contrapositive, and negation of the statement:
Converse of G: If |_a + b_| = |_a_| + |_b_|, then a - |_a_| < 1/2 and b - |_b_| < 1/2.
The converse of a statement swaps the conditions and the conclusion. To prove or disprove the converse of G, we need to provide a counterexample where |_a + b_| = |_a_| + |_b_| holds, but a - |_a_| and b - |_b_| do not satisfy the original conditions. However, we cannot find such a counterexample since the original statement has been proven to be true. Therefore, the converse is true.
Contrapositive of G: If not (a - |_a_| < 1/2 and b - |_b_| < 1/2), then not |_a + b_| = |_a_| + |_b_|.
The contrapositive of a statement negates both the conditions and the conclusion. The negation of a - |_a_| < 1/2 is a - |_a_| >= 1/2, and similarly, the negation of b - |_b_| < 1/2 is b - |_b_| >= 1/2. Therefore, the contrapositive statement becomes "If a - |_a_| >= 1/2 or b - |_b_| >= 1/2, then |_a + b_| != |_a_| + |_b_|." The contrapositive is true since it presents the logical equivalent of the original statement.
Negation of G: "There exist real numbers a and b, such that a - |_a_| < 1/2, b - |_b_| < 1/2, but |_a + b_| != |_a_| + |_b_|."
The negation of the original statement takes the form of "There exist... such that..." and negates the conditions and the conclusion. Here, we are claiming that there exist real numbers a and b that satisfy a - |_a_| < 1/2, b - |_b_| < 1/2, but |_a + b_| != |_a_| + |_b_|. To disprove this statement, we need to find a counterexample, i.e., specific values for a and b that satisfy the conditions but fail to produce the equality. However, given the proof above, we have shown that the original statement holds for all real numbers a and b. Therefore, the negation is false.