But it turns out that publications disproves the statement speak $\\sqrt2\\cdot\\sqrt2=2$ i m sorry is a reasonable number and also hence Product of two irrational number require not always be irrational. Which I uncover convincing.

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

Can who please suggest out wherein am i going dorn in my proof?

discrete-y2kcenter.orgematics logic propositional-calculus irrational-numbers
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edited jan 11 \"15 at 17:35

young name Sleziak
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asked Sep 25 \"13 at 4:48

Amit TomarAmit Tomar
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$\\begingroup$ The contrapositive is $\\neg q \\implies \\neg p$. $\\endgroup$
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Sep 25 \"13 at 4:50

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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 delinquent <Is not the product of 2 irrational numbers>. Over there is no a priori reason to expect that the delinquent <Is not the product of two irrational numbers> is indistinguishable to the assertion <Is the product of two rational numbers> (and in reality these critical two are not equivalent).

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edited Sep 25 \"13 in ~ 8:03
answered Sep 25 \"13 in ~ 5:27

DidDid
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Disprove:

Let $\\sqrt2$ it is in the irrational number. Then $\\sqrt2\\times \\sqrt2=|2|$, i beg your pardon is rational. So, the product of 2 irrational numbers is not always irrational

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answer Sep 28 \"14 in ~ 6:57

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Some more examples are

$$\\sqrt8×\\sqrt2=\\sqrt16 =4$$

$$\\sqrt2×\\sqrt32=\\sqrt64 =8$$

$$\\sqrt5×\\sqrt5=\\sqrt25 =5$$

In this method product of 2 irrational number is rational.

See more: How Many Pints To A Liter S, How Many Pints Are In A Liter

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edited Jul 25 \"15 at 18:26

Peter Woolfitt
answer Jul 25 \"15 in ~ 17:59
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