Question

Prove that the negative of an irrational number is an irrational number.

Answer 1

Answer:

Step-by-step explanation:

Yes it is irrational but lets prove it with contradiction.

The negative of any irrational number is irrational.

First, translate given statement from informal to formal language:

∀ real numbers x, if x is irrational, then −x is irrational.

Proof:

Suppose our statement is false. [we take the negation of the given statement and suppose it to be true.]

Assume, to the contrary, that

For every irrational number x such that −x is rational.

By definition of rational, we have

−x = a/b for some integers a and b with b ≠ 0. ( By zero product property )

Multiply both sides by −1, gives

x = −(a/b)

= −a/b

But −a and b are integers [since a and b are integers] and b ≠ 0 [by zero product property.] Thus, x is a ratio of the two integers −a and b with b ≠ 0. Hence, by definition of ration x is rational, which is a contradiction.

This contradiction shows that the supposition is false and so the given statement is true.

This completes the proof.

Answer 2

First, translate given statement from informal to formal language:

∀ real numbers x, if x is irrational, then −x is irrational.

Proof:

Suppose not. [we take the negation of the given statement and suppose it to be true.] Assume, to the contrary, that

∃ irrational number x such that −x is rational.

[We must deduce the contradiction.] By definition of rational, we have

−x = a/b  for some integers a and b with b ≠ 0.
Multiply both sides by −1 gives
x = -(a/b)
= −a/b
But −a and b are integers [since a and b are integers] and b ≠ 0 [by zero product property.] Thus, x is a ratio of the two integers −a and b with b ≠ 0. Hence, by definition of ration x is rational, which is a contradiction. [This contradiction shows that the supposition is false and so the given statement is true.]

This completes the proof.

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