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?
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.
1 answer
1 answer