for all real numbers x and y, if x - Prove proposition?

Question

Prove the following proposition:


1- for all real numbers x and y, if x is rational and y is irrational, then x+y is irrational.

Answer 1

Given x is rational and y is irrational, we want to prove that
(x + y) is irrational.

Let's attempt to prove this by contradiction.

Assume (x + y) is rational. It follows that

(x + y) can be expressed in the form a/b, for integers a and b and for b not equal to 0.

Therefore

x + y = a/b

If we subtract x to both sides,

y = a/b - x

Making the right hand side into a single fraction,

y = (a - xb)/b

Since x is a rational number, it can be expressed in the form c/d. Therefore, since x = c/d,

y = (a - (c/d)b)/b

Since this is a complex fraction, let's multiply top and bottom by d.

y = (ad - bc) / (bd)

Note that a, b, c, d are all integers. The integers are closed under addition, subtraction, and multiplication, so
(ad - bc) is an integer, and (bd) is an integer. This means y is a quotient of integers, and that implies y is a rational number.

However, it is given that y is an irrational number, so this is a contradiction (as y cannot be both a rational and irrational number). Therefore

(x + y) is irrational.

Answer 2

Suppose x+y is rational.
Then x+y-y = x is rational, contradiction. That shows that our supposition was wrong so x+y is irrational.



I used the fact that the difference of two rational numbers is a rational number.

Answer 3

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