Make a conjecture for each scenario
The square root of an even number
The product of two odd numbers and a multiple of 2
Find one counterexample to show that each conjecture is false.
For two real numbers a and b, a is either equal to b or greater than b.
All quardrilaterals are parallelograms.
Conjecture for the square root of an even number: The square root of an even number is always an even number.
To find a counterexample, we can try to find an even number whose square root is not even. Let's consider the number 4. The square root of 4 is 2, which is indeed an even number. Therefore, this conjecture is true.
Conjecture for the product of two odd numbers and a multiple of 2: The product of two odd numbers and a multiple of 2 is always an even number.
To find a counterexample, we need to find two odd numbers and a multiple of 2 whose product is not even. Let's consider the numbers 3, 5, and 4. The product of 3, 5, and 4 is 60, which is indeed an even number. Therefore, this conjecture is true.
Conjecture for two real numbers a and b: a is either equal to b or greater than b.
To find a counterexample, we can choose values for a and b such that a is not equal to b and not greater than b. Let's consider a = 3 and b = 5. In this case, a is not equal to b, and it is also not greater than b. Therefore, this conjecture is false.
Conjecture for all quadrilaterals are parallelograms: All quadrilaterals are parallelograms.
To find a counterexample, we need to find a quadrilateral that is not a parallelogram. Let's consider a quadrilateral with sides of lengths 3, 4, 5, and 6. This quadrilateral is not a parallelogram, as its opposite sides are not parallel. Therefore, this conjecture is false.
Conjecture 1: The square root of an even number is always an integer.
Counterexample: The square root of 6 is approximately 2.449, which is not an integer.
Conjecture 2: The product of two odd numbers and a multiple of 2 is always even.
Counterexample: Let's consider the odd numbers 3 and 5, and the multiple of 2 as 4. The product of 3, 5, and 4 is 60, which is not an even number.
Conjecture 3: For two real numbers a and b, a is either equal to b or greater than b.
Counterexample: Let's consider a = 1 and b = 3. In this case, a is less than b.
Conjecture 4: All quadrilaterals are parallelograms.
Counterexample: Consider a quadrilateral with four side lengths of 3, 4, 5, and 6. This quadrilateral is not a parallelogram.
an even number k is of the form k=2m
so, √(2m) = √2 * √m
Now the only way to get rid of the pesky √2 is for m to also have a factor of 2. That is, m=2n
√(2m) = √(2*2n) = √(4n) = 2√n
Thus, for an even number to have an integer root, it must be a multiple of 4.
What can you do on the others?