Prove the following statement by contradiction: For all real numbers r and s, if r is rational and s is irrational, then r + 2s is irrational.

Question

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The Answer is Proof (by contradiction):Suppose not. That is, suppose there are real numbers

r

and

s

such that

r

is rational and

s

is irrational and

r + 2 s

is rational. /We must show that this supposition leads logically to a contradiction.J By definition of rational,

r = \frac { a } { b } \quad

and

\quad r + 2 s = \frac { c } { d }

for some integers

a , b , c

, and

d

with

b eq 0

and

d eq 0 .

Then, by substitution,

\frac { a } { b } + 2 s = \frac { c } { d }

Solve this equation for

s

to obtain

s = \frac { 1 } { 2 } \left( \frac { c } { d } - \frac { a } { b } ight) = \frac { 1 } { 2 } \left( \frac { b c } { b d } - \frac { a d } { b d } ight) = \frac { b c - a d } { 2 b d } .

But both

b c - a d

and

2 b d

are integers because products and differences of integers are integers, and

2 b d eq 0

by the zero product property.

Hence

s

is a ratio of integers with a nonzero denominator, and so

s

is rational by definition of rational. This contradicts the supposition that

x

is irrational. [Hence the supposition is false and the given statement is true.]

Prove the Following Statement by Contradiction: for All Real Numbers