The assessment for this lesson is a discussion of a mathematical statement. You will have to determine if the statement is true or false. If it is false, you will explain your reasoning. For all real numbers a and b, 2a • b = a2 + b2
2 a ∙ b = a² + b²
Subtract 2 a ∙ b to both sides
2 a ∙ b - 2 a ∙ b = a² + b² - 2 a ∙ b
0 = a² + b² - 2 a ∙ b
a² + b² - 2 a ∙ b = 0
a² - 2 a ∙ b + b² = 0
( a - b )² = 0
Take the square of both sides
a - b = 0
Add b to both sides
a - b + b = 0 + b
a = b
The statement is true if a = b
To determine the truth value of the statement "For all real numbers a and b, 2a • b = a2 + b2," we can start by understanding the given mathematical expression.
The equation states that "2a • b = a^2 + b^2," where "•" represents some operation.
To check if the statement is true or false, we need to examine if the equation holds for all real numbers a and b. That means we need to confirm that the equation is satisfied for every possible combination of real numbers.
To proceed, let's perform the operation on both sides of the equation:
2a • b = a^2 + b^2
Multiplication has a higher priority than addition, so we need to execute the multiplication first. The equation illustrates "2a • b," which implies multiplying 2a by b. So we can rewrite the equation as:
2ab = a^2 + b^2
Now, let's consider a counterexample to prove the statement false. To disprove a universal statement like this, we only need to find one set of values for a and b that does not satisfy the equation.
For instance, let's suppose a = 2 and b = 3. Substituting these values into the modified equation, we have:
2(2)(3) = 2^2 + 3^2
12 = 4 + 9
12 ≠ 13
Since 12 does not equal 13, the equation does not hold for these particular values of a and b. Therefore, we have found a counterexample, proving that the statement is false.
To summarize, the statement "For all real numbers a and b, 2a • b = a^2 + b^2" is false. We disproved it by finding a counterexample where the equation does not hold true.