Math, asked by vyshu2171, 1 month ago

Prove that if x is rational and y is irrational, then x + y is irrational and if x 6= 0, xy is irrational.​

Answers

Answered by ankurgoswami1976
3

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.)

Answered by mithoogoswami1984
4

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.)

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