Prove irrational numbers

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we can prove that √5 is irrational through contradiction and same applies for √3.
but can we prove that √5 + √3 is irrational by contradiction?
 

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  • #2
arildno
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Show that if [tex]\sqrt{5}+\sqrt{3}[/tex] is rational, then so must [tex]\sqrt{5}-\sqrt{3}[/tex] be rational.
Derive that, hence, [tex]\sqrt{5}[/tex] is rational, a contradiction.
 
  • #3
HallsofIvy
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??? Are you saying that if x+ y is rational then x- y must be rational? What if [itex]x= \sqrt{5}[/itex] and [itex]y= -\sqrt{5}[/itex]?
 
  • #4
arildno
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HallsofIvy said:
??? Are you saying that if x+ y is rational then x- y must be rational? What if [itex]x= \sqrt{5}[/itex] and [itex]y= -\sqrt{5}[/itex]?
Not at all, but:
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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  • #5
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Is it possible for two irrational numbers to be added together to get a rational number?
 
  • #6
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Jeff Ford said:
Is it possible for two irrational numbers to be added together to get a rational number?
I don't know about that, but if you divide pi by phi you get a rational (they share a common irrational factor).
 
  • #7
arildno
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Jeff Ford said:
Is it possible for two irrational numbers to be added together to get a rational number?
Definitely, HallsofIvy gave an example of that.
 
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  • #8
Gokul43201
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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).
And what might that be ?
 
  • #9
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Gokul43201 said:
And what might that be ?
I mean phi2.

http://members.ispwest.com/r-logan/fullbook.html [Broken]
 
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  • #10
Hurkyl
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The most algorithmic approach I know for these types of problems is to find an integer polynomial with your algebraic number as a root...
 
  • #11
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).
apmcavoy said:
I mean phi2.

http://members.ispwest.com/r-logan/fullbook.html [Broken]
Are you saying that [tex]\frac{\pi}{\phi^2}\in\mathbb{Q}[/tex]?

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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  • #12
matt grime
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Jeff Ford said:
Is it possible for two irrational numbers to be added together to get a rational number?
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 like

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.
 
  • #13
HallsofIvy
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I looked at the paper referred to above. There are no proofs at all. The author refers to φ2 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.
 
  • #14
proof: if n is rational and m is irrational set [tex]n + m = l[/tex] is rational then [tex]l + (-n) = m[/tex] is rational, a contradiction
 
  • #15
CRGreathouse
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HallsofIvy said:
I looked at the paper referred to above. There are no proofs at all. The author refers to φ2 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.
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.
 
  • #16
Millie said:
we can prove that √5 is irrational through contradiction and same applies for √3.
but can we prove that √5 + √3 is irrational by contradiction?
You could also argue by contradiction that if
[tex] \sqrt5 + \sqrt3[/tex] is rational then so must [tex](\sqrt5 + \sqrt3)^2[/tex] and something bad will happen.
 
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  • #17
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we can prove that √5 is irrational through contradiction and same applies for √3.
but can we prove that √5 + √3 is irrational by contradiction?
show me how by contradiction
 
  • #18
uart
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show me how by contradiction
Assume [itex]\sqrt{5} = \frac{a}{b}[/itex] for some integers "a" and "b".

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.
 
  • #19
show me how by contradiction
The contradiction arises when you try to proof that sqrt(15) is rational.
 

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