But it turns out that books disproves the statement saying $sqrt2cdotsqrt2=2$ which is a rational number and hence Product of two irrational number need not always be irrational. Which I find convincing.

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

Can someone please point out where am I going wrong in my proof?


discrete-sdrta.netematics logic propositional-calculus irrational-numbers
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Folshort
edited Jan 11 "15 at 17:35
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Martin Sleziak
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asked Sep 25 "13 at 4:48
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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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The negation of the assertion <Is the product of two irrational numbers> is the assertion <Is not the product of two irrational numbers>. There is no a priori reason to expect that the assertion <Is not the product of two irrational numbers> is equivalent to the assertion <Is the product of two rational numbers> (and in fact these last two are not equivalent).


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Folshort
edited Sep 25 "13 at 8:03
answered Sep 25 "13 at 5:27
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DidDid
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Disprove:

Let $sqrt2$ be the irrational number. Then $sqrt2 imes sqrt2=|2|$, which is rational. So, the product of two irrational numbers is not always irrational


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answered Sep 28 "14 at 6:57
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Hassan MuhammadHassan Muhammad
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Some more examples are

$$sqrt8×sqrt2=sqrt16 =4$$

$$sqrt2×sqrt32=sqrt64 =8$$

$$sqrt5×sqrt5=sqrt25 =5$$

In this way product of two irrational number is rational.

See more: Is That The Oxidation Number Of N In Nano3 Is _____? Assignment Of Oxidation Numbers


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Folshort
edited Jul 25 "15 at 18:26
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Peter Woolfitt
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answered Jul 25 "15 at 17:59
shrutishruti
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