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.

Question

Dalton Coleman

Mathematics

2 July, 08:20

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.

+5

Answers (1)

Know the Answer?

Jase Gutierrez · Answer 1

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