But it transforms out that publications disproves the statement speak $sqrt2cdotsqrt2=2$ i m sorry is a reasonable number and hence Product of 2 irrational number need not constantly be irrational. I m sorry I discover convincing.

You are watching: The product of two irrational numbers is rational

Can someone please allude out wherein am ns going not correct in mine proof?

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edited january 11 "15 in ~ 17:35

boy name Sleziak

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Amit TomarAmit Tomar

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$egingroup$ The contrapositive is $ eg q implies eg p$. $endgroup$

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## 3 answers 3

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The negation of the assertion <

*Is the product of 2 irrational numbers*> is the assertion <

*Is no the product of 2 irrational numbers*>. There is no a priori factor to expect that the delinquent <

*Is no the product of 2 irrational numbers*> is indistinguishable to the assertion <

*Is the product of two rational numbers*> (and in fact these critical two room not equivalent).

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edited Sep 25 "13 in ~ 8:03

answered Sep 25 "13 at 5:27

DidDid

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Disprove:

Let $sqrt2$ be the irrational number. Climate $sqrt2 imes sqrt2=|2|$, i beg your pardon is rational. So, the product of two irrational number is not always irrational

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answer Sep 28 "14 in ~ 6:57

Hassan MuhammadHassan Muhammad

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Some an ext examples room

$$sqrt8×sqrt2=sqrt16 =4$$

$$sqrt2×sqrt32=sqrt64 =8$$

$$sqrt5×sqrt5=sqrt25 =5$$

In this means product of 2 irrational number is rational.

See more: What Are The Factors Of 361 ? What Are All The Factors Of 361

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edited Jul 25 "15 in ~ 18:26

Peter Woolfitt

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