Try to prove the hypothesis from part b that the sum of a rational and an irrational number is an irrational number. let’s say is x an irrational number and y is a rational number. the rational number y can be written as y=a/b , where a and b are integers and b does not equal 0. leave the irrational number x as x since it can’t be written as the ratio of two integers. prove the hypothesis using a proof by contradiction. in other words, try to show that x+y equals a rational number instead of an irrational number. let the sum equal m/n , where m and n are integers and n does not equal 0. plug in the values to get this equation: x+a/b=m/n now, solve the equation for x .

The next step of your proof is to subtract (a/b) from both sides.

Then you get, x = (m/n) - (a/b)

Since rationals are closed over addition, (m/n) + (-a/b) is a rational number.

Therefore, x (an irrational number) = a rational number                            This is a false statement which is a contradiction. So, the assumption was incorrect.

Thus, the sum of a rational and irrational number is an irrational number. QED

0.16 or 16%

step-by-step explanation:

lets take a coin for example.

if you toss it twice, it has a 1/4 chance of landing on heads both times because 1/2 * 1/2 is 1/4.

with the cards, the chance of picking dog is 0.4. having two people pick one is like picking a card twice in a row. so 0.4 * 0.4 is 0.16, or 16%