Final answer:
To demonstrate that if x and y are real numbers with x < y, then x < (x + y)/2 < y, we add x and y to different sides of the initial inequality, divide by 2, and combine the results to obtain the required expression.
Step-by-step explanation:
To prove that if x and y are real numbers such that x < y, then x < (x + y)/2 < y, we can use the properties of inequalities to manipulate the expressions.
First, since x < y, adding x to both sides of the inequality gives us 2x < x + y. Dividing both sides by 2, which is positive and does not change the direction of the inequality, we get x < (x + y)/2.
Similarly, adding y to both sides of the initial inequality x < y, we get x + y < 2y. Again, dividing by 2 gives us (x + y)/2 < y.
Combining both results, we have x < (x + y)/2 < y, which is what we wanted to prove.
To prove that if x and y are real numbers such that x < y, then x < (x + y)/2 < y, we can start by considering the midpoint of x and y, which is (x + y)/2. Since x < y, adding x and y must produce a value larger than 2x. Similarly, adding x and y must produce a value smaller than 2y. Therefore, x < (x + y)/2 < y.