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but can we prove that √5 + √3 is irrational by contradiction?

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but can we prove that √5 + √3 is irrational by contradiction?

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arildno

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Derive that, hence, [tex]\sqrt{5}[/tex] is rational, a contradiction.

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HallsofIvy

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arildno

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Not at all, but:HallsofIvy said:

Let [tex]\sqrt{5}+\sqrt{3}=p[/tex] where p is rational.

We therefore have:

[tex]\frac{2}{\sqrt{5}-\sqrt{3}}=p[/tex], that is:

[tex]\sqrt{5}-\sqrt{3}=m[/tex], m=2/p rational.

That is, since the difference of squares of [tex]\sqrt{5}[/tex] and [tex]\sqrt{3}[/tex] is rational, and [tex]\sqrt{5}+\sqrt{3},\sqrt{5}-\sqrt{3}\neq{0}[/tex] my argument holds.

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Is it possible for two irrational numbers to be added together to get a rational number?

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I don't know about that, but if you divide pi by phi you get a rational (they share a common irrational factor).Jeff Ford said:Is it possible for two irrational numbers to be added together to get a rational number?

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arildno

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Definitely, HallsofIvy gave an example of that.Jeff Ford said:Is it possible for two irrational numbers to be added together to get a rational number?

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Gokul43201

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And what might that be ?apmcavoy said:I don't know about that, but if you divide pi by phi you get a rational (they share a common irrational factor).

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I mean phiGokul43201 said:And what might that be ?

http://members.ispwest.com/r-logan/fullbook.html [Broken]

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Hurkyl

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CRGreathouse

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apmcavoy said:I don't know about that, but if you divide pi by phi you get a rational (they share a common irrational factor).

Are you saying that [tex]\frac{\pi}{\phi^2}\in\mathbb{Q}[/tex]?apmcavoy said:I mean phi^{2}.

http://members.ispwest.com/r-logan/fullbook.html [Broken]

Since [tex]\phi=\frac{1+\sqrt5}{2}[/tex] is algebraic and algebraic numbers are closed under multiplication and division, that would mean that pi is algebraic (all rationals are algebraic). The paper you link to, though, admits that "[tex]\pi[/tex] is a transcendental number" and that trancendentals aren't algebraic (by definition).

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matt grime

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yes, as has been shown, but more gnerally (I can't recall where i first read this wordy description) but if you are trying to decide if two nasties can make a nice then try thinking if a nice minus a nasty can be nasty. here, obviously the rational minus irrational must be irrational, so anything likeJeff Ford said:Is it possible for two irrational numbers to be added together to get a rational number?

1-sqrt(2) and sqrt(2) add together to give what you want.

the same works for continuous and discontinuous functions, invertible and uninvertible matrices, rational and irrational numbers, algebraic and transcendental numbers and so on.

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HallsofIvy

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CRGreathouse

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Yes, that was what most bothered me. Normally there's a certain semblance of internal consistency in these pseudoproofs, but this one just didn't define enough to come to a contradiction that I was patient enough to find.HallsofIvy said:I looked at the paper referred to above. There are no proofs at all. The author refers to φ^{2}as a "composite" number with no explanation of what "composite" is supposed to mean for a non-integer number. All he does is show that 15 decimal place computations come out the same, then spends the rest of the paper defining "fraction", "rational", "irrational", etc.

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You could also argue by contradiction that ifMillie said:

but can we prove that √5 + √3 is irrational by contradiction?

[tex] \sqrt5 + \sqrt3[/tex] is rational then so must [tex](\sqrt5 + \sqrt3)^2[/tex] and something bad will happen.

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show me how by contradiction

but can we prove that √5 + √3 is irrational by contradiction?

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uart

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Assume [itex]\sqrt{5} = \frac{a}{b}[/itex] for some integers "a" and "b".show me how by contradiction

Then

[tex]a^2 = 5 b^2[/tex]

The LHS of the above equation has prime factor "5" to an even power while the RHS has prime factor "5" to an odd power. This is an obvious contradiction.

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The contradiction arises when you try to proof that sqrt(15) is rational.show me how by contradiction

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