Proving a+br is Irrational Using Proof by Contradiction

In summary, you can use the fact that the set of rational numbers is closed under addition and multiplication to show that r is a quotient of integers.
  • #1
mr_coffee
1,629
1
Hello everyone! THe directions are the following:
Carefully formulate the negations of each of the statements. Then prove each statement by contradiction.

Here is the problem:
If a and b are rational numbers, b!=0, and r is an irrational number, then a+br is irrational.

I made it into a universal statement:
[tex]\forall[/tex] real numbers a and b and r, if a nd b are rational such that b != 0, and r is irrational, then a + br is rational.

I then took the negation:
[tex]\exists[/tex] rational numbers a and b, b != 0 and irrational number r such that a + br is rational.

Proof:
Suppose not. That is, suppose there are rational numbers a and b, with b != 0 and r is an irrational number such that a + br is rational. By definition of rational a + br = e/f and a = m/n and b = x/y for some integers e, f, m, n, x, and y with n,y,x, and f != 0. By substitution

a + br = e/f
a = m/n;
r = irrational
b = x/y

m/n + (x/y)*r = e/f

r = y/x(e/f – m/n)

r = (yen-fmy)/fnx

now yen-fmy and fnx are both integers, and fnx != 0. Hence r is a quotient of the integers yen-fmy and fnx with fnx != 0. Thus, by definition of rational, r is rational, which contradicts the supposition that a+br is irrational.


I know there are a lot of variables, but i wasn't sure how else to show that r is infact a quotient of integers.

Any comments on what i did if its correct or flawed? Thank you!
 
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  • #2
What you've done is perfectly correct but you are right- it's hard to read with all those letters.

You might just use the fact that the set of rational numbers is closed under addition and multiplication- and every non-zero rational number has a rational multiplicative inverse.

If a+ br= c then r= (c-a)(1/b). If a, b, c, b not 0, are all rational then c-a is rational (because the rational numbers are closed under addition), 1/b is rational (b is not 0 so has a rational multiplicative inverse), (c-a)(1/b) is rational (the rational numbers are closed under multiplication).
 
  • #3
That does look a lot nicer, thanks for the help! :biggrin:
 

1. What is "proof by contradiction"?

Proof by contradiction is a method of mathematical proof in which a statement is proved to be true by assuming its opposite (or negation) to be true and then showing that this leads to a contradiction. This contradiction proves that the original statement must be true.

2. Why is proof by contradiction used to prove irrationality?

Proof by contradiction is often used to prove irrationality because it is a powerful technique that can be used to show that a number cannot be expressed as a ratio of two integers. This is a key characteristic of irrational numbers.

3. What is the statement being proven in this proof?

The statement being proven is that the number a+br, where a and b are rational numbers and r is an irrational number, is also irrational.

4. What is the general process for proving a+br is irrational using proof by contradiction?

The general process for proving a+br is irrational using proof by contradiction involves assuming that a+br is rational and can be expressed as a ratio of two integers, then manipulating the equation to show that this leads to a contradiction. This contradiction proves that a+br cannot be rational and is therefore irrational.

5. Can this proof be used for any combination of rational and irrational numbers?

Yes, this proof can be used for any combination of rational and irrational numbers. As long as the irrational number is being added to or multiplied by a rational number, this proof can be applied to prove its irrationality.

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