Prove that of sum square root of 2 and square root of 3 is not rational

In summary, a rational number is a number that can be expressed as the quotient or fraction of two integers. To prove that the sum of \(\sqrt{2}\) and \(\sqrt{3}\) is not rational, we use a proof by contradiction and show that the assumption that \(\sqrt{2} + \sqrt{3}\) is rational leads to a contradiction. This proof is significant in demonstrating the irrationality of a specific mathematical expression and is important in understanding the properties of irrational numbers.
  • #1
transgalactic
1,395
0
prove that the square root of 2 plus the square root of 3 is not rational?

does always the sum of two not rational numbers is a not rational number?



i know the proof 2 = a^2/b^2
i separately proved that square root of 2 and square root of 3 are irrational

how two prove that the sum of two such numbers is irrational too?

whats the formal equation proof?
 
Last edited:
Physics news on Phys.org
  • #2
transgalactic said:
prove that the square root of 2 plus the square root of 3 is not rational?

does always the sum of two not rational numbers is a not rational number?



i know the proof 2 = a^2/b^2
i separately proved that square root of 2 and square root of 3 are irrational

how two prove that the sum of two such numbers is irrational too?

whats the formal equation proof?

Try a proof by contradiction. Suppose that [tex]\sqrt{2} + \sqrt{3} = \frac{a}{b}[/tex]
where a and b are integers with no common factors.

For your second question [tex]\sqrt{5} + (-\sqrt{5}) = 0 [/tex]
Those are both irrational numbers, but their sum is rational.
 
  • #3
Well clearly two irrational numbers can sum to give a rational number. For example (2+sqrt(2)) + (2 -sqrt(2)).

To show that the sqrt(2) + sqrt(3) is irrational, you can start with
a = sqrt(2) + sqrt(3)
Manipulate to get
a^4 - 10a^2 + 1 = 0

Then you can use the rational roots theorem, that shows that the only rational roots of this equation can be +- 1. http://en.wikipedia.org/wiki/Rational_root_theorem
 
  • #4
Mark44 said:
For your second question [tex]\sqrt{5} + (-\sqrt{5}) = 0 [/tex]
Those are both irrational numbers, but their sum is rational.

Somehow, that just seems like cheating, but it does fit the problem we were given.
 
  • #5
Chaos2009 said:
Somehow, that just seems like cheating, but it does fit the problem we were given.

Any mathematical statement can be disproved by a counterexample. The simpler the counterexample, the better.
 
  • #6
ok i will try to prove by contradiction:
suppose
(2)^0.5 + (3)^0.5 is a rational number
(if we multiply a rational number by a rational number we will get a rational number "h")
5+2*(2)^0.5 * (3)^0.5=h
24=h^2 -10*h +25

h^2 -10*h +1=0
what to do now?
 
  • #7
As I suggested earlier, make use of the rational roots theorem.
 
  • #8
by this rational roots theorem
the possible roots is +1 and -1

not one of the represent the actual roots of h^2 -10*h +1=0

what is the next step in the prove?
 
  • #10
You don't need anything but parity to prove this.

Prove



[tex]\sqrt{2}+\sqrt{3} = (p/q)[/tex]
[tex]2 \sqrt{6}+5 = p^{2}/q^{2}[/tex]

You know that the rationals form a field, which implies all the terms in the sum must be a rational.

Now all that remains is to show [tex]\sqrt{6}[/tex] is irrational. So just show that if it equals p/q, p and q must both be even.
 
Last edited by a moderator:
  • #11
transgalactic said:
ok i will try to prove by contradiction:
suppose
(2)^0.5 + (3)^0.5 is a rational number
(if we multiply a rational number by a rational number we will get a rational number "h")
5+2*(2)^0.5 * (3)^0.5=h
24=h^2 -10*h +25

h^2 -10*h +1=0
what to do now?

the root test showed me that there is no rational roots for this equation.
is it fair to consider "h" irrational because of it?

i was told that if we don't have a rational roots it doesn't mean that
"h" is always irrational,"h" could be a complex number.

??
 
  • #12
If a number is complex, it certainly isn't a rational number. If h^2 - 10h + 1 = 0 doesn't have any rational roots, then there is no solution for h that is rational.
 
Last edited:
  • #13
you meant "irrational" (in the end) right?
 
  • #14
I edited my reply. The last part should have said "there is no solution that is rational."
 
  • #15
nicksauce said:
Well clearly two irrational numbers can sum to give a rational number. For example (2+sqrt(2)) + (2 -sqrt(2)).

To show that the sqrt(2) + sqrt(3) is irrational, you can start with
a = sqrt(2) + sqrt(3)
Manipulate to get
a^4 - 10a^2 + 1 = 0

Then you can use the rational roots theorem, that shows that the only rational roots of this equation can be +- 1. http://en.wikipedia.org/wiki/Rational_root_theorem

how did you arrive at a^4 - 10a^2 + 1 = 0?
need help
 
  • #16
Well since we've already summoned this thread from the grave...
jay17 said:
how did you arrive at a^4 - 10a^2 + 1 = 0?
need help

[tex]a=\sqrt{2}+\sqrt{3}[/tex]

[tex]a^2=(\sqrt{2}+\sqrt{3})^2=2+2\sqrt{2}\sqrt{3}+3=5+2\sqrt{6}[/tex]

[tex](a^2-5)^2=(2\sqrt{6})^2[/tex]

[tex]a^4-10a^2+25=24[/tex]

[tex]a^4-10a^2+1=0[/tex]
 

What is a Rational Number?

A rational number is a number that can be expressed as the quotient or fraction of two integers, where the denominator is not zero. In other words, a number \(x\) is rational if it can be written in the form \(x = \frac{a}{b}\), where \(a\) and \(b\) are integers and \(b\) is not equal to zero.

How can we prove that the sum of \(\sqrt{2}\) and \(\sqrt{3}\) is not rational?

To prove that the sum of \(\sqrt{2}\) and \(\sqrt{3}\) is not a rational number, we will use a proof by contradiction. We will assume that the sum \(\sqrt{2} + \sqrt{3}\) is rational and then show that this assumption leads to a contradiction.

  1. Assume that \(\sqrt{2} + \sqrt{3}\) is rational: Let's assume that \(\sqrt{2} + \sqrt{3}\) is a rational number, which means it can be expressed as \(\sqrt{2} + \sqrt{3} = \frac{a}{b}\), where \(a\) and \(b\) are integers with no common factors other than 1 (\(a\) and \(b\) are coprime).
  2. Square both sides of the equation: Square both sides of the equation \(\sqrt{2} + \sqrt{3} = \frac{a}{b}\) to eliminate the square roots: \[(\sqrt{2} + \sqrt{3})^2 = \left(\frac{a}{b}\right)^2\]
  3. Expand and simplify the left side: On the left side, use the distributive property and simplify: \[(\sqrt{2} + \sqrt{3})^2 = 2 + 2\sqrt{6} + 3 = 5 + 2\sqrt{6}\]
  4. Write the right side as a fraction: On the right side, square \(\frac{a}{b}\) to get \(\left(\frac{a}{b}\right)^2 = \frac{a^2}{b^2}\).
  5. Equate the expressions: Equate the left side and the right side: \[5 + 2\sqrt{6} = \frac{a^2}{b^2}\]
  6. Isolate the radical term: Isolate the term with the square root on the left side: \[2\sqrt{6} = \frac{a^2}{b^2} - 5\]
  7. Square both sides again: Square both sides of the equation to eliminate the square root: \[(2\sqrt{6})^2 = \left(\frac{a^2}{b^2} - 5\right)^2\]
  8. Simplify the left side: On the left side, simplify \(2\sqrt{6} \cdot 2\sqrt{6} = 24\).
  9. Expand and simplify the right side: On the right side, expand and simplify \(\left(\frac{a^2}{b^2} - 5\right)^2\).
  10. Arrive at a contradiction: The equation will lead to a contradiction because the left side is 24 (an even number), and the right side will not be an even number since \(a\) and \(b\) are coprime, and \(a^2\) and \(b^2\) do not share any common factors. This contradiction arises from the initial assumption that \(\sqrt{2} + \sqrt{3}\) is rational.

Since the assumption that \(\sqrt{2} + \sqrt{3}\) is rational leads to a contradiction, we conclude that \(\sqrt{2} + \sqrt{3}\) is not a rational number. It is an irrational number.

Why is this proof important?

This proof is significant because it demonstrates the irrationality of a specific mathematical expression, \(\sqrt{2} + \sqrt{3}\), using a proof by contradiction. Understanding the properties of irrational numbers and how to prove them is fundamental in mathematics and has applications in various mathematical fields, including number theory and algebra.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
806
  • Calculus and Beyond Homework Help
Replies
1
Views
180
  • Precalculus Mathematics Homework Help
Replies
3
Views
789
Replies
23
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
734
  • Calculus and Beyond Homework Help
Replies
31
Views
2K
  • General Math
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
9K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
524
Back
Top