Prove that if x is rational and y is irrational, then x + y is irrational and if x 6= 0, xy is irrational.
Answers
Answer:
Hint: Consider (x+y) - x
Explaination:
As is very often the case, we do not need to write this as a proof by contradiction. We can prove the contrapositive directly.
We can prove directly:
x is rational ⇒ ( x+y is rational ⇒ y is rational)
(using a, b∈Q⇒a−b∈Q -- that is, Q is closed under subtraction)
Therefore (by contraposition of the imbedded conditional)
x is rational ⇒ ( y is not rational ⇒x+y is not rational)
This is logically equivalent to
( x is rational & y is not rational)
⇒x+y is not rational)
By contradiction
Suppose p and ¬q and r .
Prove a contradiction and conclude that if p and ¬q , then ¬r .
By contrapositive
Suppose p and ¬r . Prove that q.
Conclude that If p and ¬q , then r .
The two methods are very closely related and I don't know of anyone who accepts one and not the other. (Although many/most/all intuitionists refuse to accept either contradiction or contrapositive.)
Proof by contrapositive
Suppose that x is rational and x+y is rational.
Then the difference (x+y)−x=y is rational.
Hence is we know that y is irrational, then x+y must have been irrational. (Otherwise, y would have been rational after all.)
Answer:
Hint: Consider
(x+y)−x
Explanation:
As is very often the case, we do not need to write this as a proof by contradiction. We can prove the contrapositive directly.
We can prove directly:
x is rational ⇒ ( x+y is rational ⇒y is rational)
(using a,b∈Q⇒a−b∈Q -- that is, Q is closed under subtraction)
Therefore (by contraposition of the imbedded conditional)
x is rational ⇒ ( y is not rational ⇒x+y is not rational)
This is logically equivalent to
( x is rational & y is not rational)
⇒ x+y is not rational)
By contradiction
Suppose p and ¬q and r .
Prove a contradiction and conclude that if p and ¬q , then ¬r .
By contrapositive
Suppose p and ¬r .
Prove that q .
Conclude that If p and ¬q , then r .
The two methods are very closely related and I don't know of anyone who accepts one and not the other. (Although many/most/all intuitionists refuse to accept either contradiction or contrapositive.)
Proof by contrapositive
Suppose that x is rational and x+y is rational.
Then the difference (x+y)−x=y is rational.
Hence is we know that y is irrational, then x+y must have been irrational. (Otherwise, y would have been rational after all.)