Proof: x is irrational => sqrt(x) is irrational

In summary, the conversation discusses how to prove the statement "If x is irrational, then √x is irrational" and the use of the contrapositive in making the proof easier. It also clarifies the difference between proving by contrapositive and by contradiction.
  • #1
mattmns
1,128
6
Ok so I am to prove: If x is irrational, then [tex]\sqrt{x}[/tex] is irrational. So I started by trying to prove the contrapositive: If [tex]\sqrt{x}[/tex] is rational, then x is rational.

So then [tex]\sqrt{x} = \frac{m}{n}[/tex] For integers m and n, n[tex]\neq[/tex]0

Then square both sides. [tex]x = \frac{m^2}{n^2}[/tex]

This is clearly rational because m^2 and n^2 are integers.

Now, is this a satisfactory proof? I am sure it is, it just seems as though it was too easy. Did my teacher ask it because it shows how proving the contrapositive can sometimes make life easy? Thanks.
 
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  • #2
Yes, that's correct, and it is an easy proof. It boils down to this: Every square of a rational number is rational, and so these rational squares are the only numbers which have rational square roots.
 
  • #3
Did my teacher ask it because it shows how proving the contrapositive can sometimes make life easy? Thanks.

Yep! Ideally, when you're posed with the problem of proving a statement, looking at its contrapositive should become second nature! With luck, you'll get to the point where you barely even notice the difference between A→B and ~B→~A!
 
  • #4
Actually that is incorrect.
The negation of "irrational" is simply "not irrational". For a number to be "not irrational" has 2 cases. The number must be either complex (including i) or rational. Thus your statement of what the contrapositive is is not logically equivalent. This proof must be done by contradiction not by contrapositive.
 
  • #5


Yes, your proof is correct and well-reasoned. It is a good example of how proving the contrapositive can make a proof easier. By assuming the negation of the statement and showing that it leads to a contradiction, you have effectively proven the original statement. It is also a good demonstration of how understanding the properties and definitions of irrational and rational numbers can lead to a proof. Well done.
 

FAQ: Proof: x is irrational => sqrt(x) is irrational

1. What does it mean for a number to be irrational?

A number is considered irrational if it cannot be expressed as a fraction of two integers (whole numbers). In other words, it cannot be written in the form a/b where a and b are integers and b is not equal to 0. Irrational numbers have non-repeating and non-terminating decimal representations.

2. How do you prove that a number is irrational?

There are several methods to prove that a number is irrational. One method is to assume that the number is rational and then use proof by contradiction to show that this assumption leads to a contradiction. Another method is to use the rational root theorem, which states that if a polynomial equation with integer coefficients has a rational root, then that root must be an integer.

3. How does the proof for x being irrational imply that sqrt(x) is irrational?

The proof for x being irrational is based on the assumption that x is rational and leads to a contradiction. Since the statement "x is rational" is false, the negation of this statement, which is "x is irrational", must be true. If x is irrational, then its square root, sqrt(x), must also be irrational because sqrt(x) cannot be expressed as a fraction of two integers, just like x.

4. Can a number be both rational and irrational?

No, a number cannot be both rational and irrational. By definition, a number is either rational or irrational. A number that can be expressed as a fraction of two integers is rational, while a number that cannot be expressed as a fraction of two integers is irrational.

5. Are there any other implications of the proof for x being irrational?

Yes, there are other implications of the proof for x being irrational. For example, the proof can be extended to show that if x is rational, then any root of x (such as the cube root or fourth root) is also rational. Additionally, the proof can be used to show that if x is irrational, then any power of x (such as x^2 or x^3) is also irrational.

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