Is -2sqrt2 an irrational number ?

danielvnl said:
I think yes.
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\(\displaystyle Assume\ there\ exist\ one\ or\ more\ pairs\ of\ positive\ integers\ a_i\ and\ b_i\ such\ that\ \dfrac{a_i}{b_i} = \sqrt{2}\ and\ a_1 \le a_i.\)

\(\displaystyle \dfrac{a_1}{b_1} = \sqrt{2} \implies \left(\dfrac{a_1}{b_1}\right)^2 = 2 \implies \dfrac{a_1^2}{b_1^2} = 2 \implies a_1^2 = 2b_1^2 \implies\)

\(\displaystyle a_1\ is\ even \implies there\ exists\ positive\ integer\ c\ such\ that\ 2c = a_1 \implies c < a\ and\ a_1^2 = 4c^2 \implies 2b_1^2 = a_1^2 = 4c^2\implies\)

\(\displaystyle b^2 = 2c^2 \implies b\ is\ even \implies there\ exists\ positive\ integer\ d\ such\ that\ 2d = b_1 \implies \)

\(\displaystyle \sqrt{2} = \dfrac{a_1}{b_1} = \dfrac{2c}{2d} = \dfrac{c}{d} \implies\)

\(\displaystyle there\ exists\ integer\ k\ such\ that\ c = a_k\ and\ d = b_k \implies a_k < a_1,\ a\ contradiction.\)

Because the assumption leads to a contradiction, the assumption is false.

This proof is ancient, well over 2000 years old.
 
danielvnl said:
[h=2]Is -2sqrt2 an irrational number ?[/h]I think yes.
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using the proof above (although Jeff is not 2000 years old):

-2√2 = -2 * (irrational number) = (rational number) * (irrational number) = irrational number.

Remember, however, (irrational number) * (irrational number) can produce (rational number) - e.g. → √2 * √2 = 2
 
Subhotosh Khan said:
using the proof above (although Jeff is not 2000 years old):

-2√2 = -2 * (irrational number) = (rational number) * (irrational number) = irrational number.

Remember, however, (irrational number) * (irrational number) can produce (rational number) - e.g. → √2 * √2 = 2
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Subhotosh Khan is quite right. My proof was incomplete. I shall go visit the corner where denis lives, but Subhotosh must go too because he forgot that there is one rational number that always gives a rational product when multiplied by an irrational number.

\(\displaystyle Assume\ u\ is\ an\ irrational\ number,\ a\ and\ b\ are\ non\text{-}zero\ integers,\ and\ u * \dfrac{a}{b}\ is\ rational \implies\)

\(\displaystyle there\ exist\ integers\ c\ and\ d\ such\ that\ u * \dfrac{a}{b} = \dfrac{c}{d} \implies u = \dfrac{bc}{ad},\ but\)

\(\displaystyle a,\ b,\ c,\ and\ d\ are\ integers \implies bc\ and\ ad\ are\ integers \implies\)

\(\displaystyle u\ is\ a\ rational\ number,\ which\ is\ a\ contradiction.\)

So an irrational number times a non-zero rational number is always irrational.

Come Subhotosh, we shall go to the corner together. I am so old that I need an arm to lean upon.

Edit: Halls of Ivy's proof is much more elegant them mine.
 
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The difference is that the set of rational numbers is closed under multiplication. If \(\displaystyle -2\sqrt{2}\) were rational, then multiplying it by the rational number, -1/2, should give a rational number. But \(\displaystyle (-1/2)(2\sqrt{2})= \sqrt{2}\) which is NOT rational.

The set of irrational numbers is not closed under multiplication.
 
So the afirmative

"The addition of a irrational number to another irrational number "(irrational number)+(irrational number)" is always a irrational number"

Is false ???

Can you answer this question ???
 
danielvnl said:
"The addition of a irrational number to another irrational number "(irrational number)+(irrational number)" is always a irrational number" Is false ???
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What is \(\displaystyle (\sqrt{2})+(-\sqrt{2})=~?\)
 
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