sum and product of rational and irrational numbers

[logistic_guy]

Prove or disprove each of the following statements.

a) The sum of a rational and an irrational number is irrational.
b) The sum of two irrational numbers is irrational.
c) The product of a rational number and an irrational number is irrational.
d) The product of two irrational numbers is irrational.

 

[khansaheb]

Prove or disprove each of the following statements.

a) The sum of a rational and an irrational number is irrational.
b) The sum of two irrational numbers is irrational.
c) The product of a rational number and an irrational number is irrational.
d) The product of two irrational numbers is irrational.

show us your effort/s to solve this problem.

 

[logistic_guy]

Let me try to solve (a).

I'll start with the definition of rational numbers: \(\displaystyle \mathbb{Q} = \left\{\frac{a}{b} \ \big| \ a, b \in \mathbb{Z}, b \neq 0 \right\}\)

*The product, the sum, or the difference of two integers is also an integer.

I will let \(\displaystyle x\) be a rational number. This means that \(\displaystyle x = \frac{a}{b}, \ \ b \neq 0\). And I will let \(\displaystyle y\) be an irrational number.

I will assume that \(\displaystyle x + y\) is a rational number. This means that \(\displaystyle x + y = \frac{c}{d}, \ \ d \neq 0\).

Then

\(\displaystyle x + y = \frac{a}{b} + y = \frac{c}{d}\)

\(\displaystyle y = \frac{c}{d} - \frac{a}{b} = \frac{cb}{db} - \frac{ad}{bd} = \frac{cb - ad}{bd}, \ \ bd \neq 0\).

This shows that \(\displaystyle y\) is a rational number which is a contradiction. Then, \(\displaystyle x + y\) is an irrational number and the statement in \(\displaystyle (a)\) is true.

I will continue in the next post.

 

[fresh_42]

You can use this idea directly on rational numbers, not only integers. The rational numbers are closed under addition and negation. [imath] r+i=r' [/imath] implies [imath] i=r'-r=r'+(-r) [/imath] provides the contradiction.

 

[logistic_guy]

You can use this idea directly on rational numbers, not only integers.

Of course as I previously in other thread proved that the sum and product of two rational numbers are rational.

[imath] r+i=r' [/imath] implies [imath] i=r'-r=r'+(-r) [/imath] provides the contradiction.

This is beautiful, but I prefer to not use the idea directly to practise some definitions.

Thanks professor.

 

[fresh_42]

This is beautiful, but I prefer to not use the idea directly to practise some definitions.

I think my argument applies the definitions and uses the essential properties rather than calculations that hide the crucial points.

Let [imath] r\in \mathbb{Q} [/imath] and [imath] i \in \mathbb{R}\setminus \mathbb{Q}. [/imath]
Preparation of an inductive conclusion.

Assume [imath] r':=r+i \in \mathbb{Q}. [/imath]
Preparation of reduction ad absurdum.

Then
[math]\begin{array}{lll} i &=\underbrace{r' + \underbrace{(-r)}_{\in \mathbb{Q}^(1)}}_{\in \mathbb{Q}^(2)} \in \mathbb{Q} \end{array}[/math]where (1) is closure under negation and (2) is closure under addition.

We arrived ad absurdum since [imath] i\in \mathbb{Q}\cap \left(\mathbb{R}\setminus \mathbb{Q}\right)=\emptyset [/imath] cotradicting our assumption, and proving [imath] i\not\in \mathbb{Q}. [/imath] Finally, inductive conclusion allows us to claim that this is true for any choices of [imath] r [/imath] and [imath] i [/imath] proving the general statement (a).

Writing it that way explicitly displays all the principles we had to use.

 

[logistic_guy]

I think my argument applies the definitions and uses the essential properties rather than calculations that hide the crucial points.

Let [imath] r\in \mathbb{Q} [/imath] and [imath] i \in \mathbb{R}\setminus \mathbb{Q}. [/imath]
Preparation of an inductive conclusion.

Assume [imath] r':=r+i \in \mathbb{Q}. [/imath]
Preparation of reduction ad absurdum.

Then
[math]\begin{array}{lll} i &=\underbrace{r' + \underbrace{(-r)}_{\in \mathbb{Q}^(1)}}_{\in \mathbb{Q}^(2)} \in \mathbb{Q} \end{array}[/math]where (1) is closure under negation and (2) is closure under addition.

We arrived ad absurdum since [imath] i\in \mathbb{Q}\cap \left(\mathbb{R}\setminus \mathbb{Q}\right)=\emptyset [/imath] cotradicting our assumption, and proving [imath] i\not\in \mathbb{Q}. [/imath] Finally, inductive conclusion allows us to claim that this is true for any choices of [imath] r [/imath] and [imath] i [/imath] proving the general statement (a).

Writing it that way explicitly displays all the principles we had to use.

I understand your explanation, but I have never used induction if that what meant by the inductive conclusion.

rather than calculations that hide the crucial points.

How do my calculations hide the crucial points? I have started with a definition and shown every important part clearly till I concluded that \(\displaystyle x + y\) is irrational. If you think that my proof hide the crucial points, do you mean that it is not complete?

 

[fresh_42]

I understand your explanation, but I have never used induction if that what meant by the inductive conclusion.

Not induction, inductive conclusion. It is the conclusion from the special case to the general case. This is allowed as long as we do not impose any restrictions on the special case. It is the opposite of a deductive conclusion where we conclude from the general case to the special case.

Deductive: The sum of a rational number and an irrational number is irrational, hence [imath] 1+\sqrt{2} [/imath] is irrational.

Inductive: Let [imath] r [/imath] be any rational number and [imath] i [/imath] any irrational number. Then ... [imath] r+i [/imath] is irrational and all sums of a rational number and an irrational number are irrational, hence [imath] 1+\sqrt{2} [/imath] is irrational.

How do my calculations hide the crucial points? I have started with a definition and shown every important part clearly till I concluded that \(\displaystyle x + y\) is irrational. If you think that my proof hide the crucial points, do you mean that it is not complete?

It is the same as my proof, just with more words, an unnecessary restriction to integers, and a proposed second part for rationals. Maybe I should have said distract instead of hide. I called it closure under negation and addition, and you wrote [imath] \dfrac{a}{b}\, , \,\dfrac{c}{d}\in \mathbb{Q} \Longrightarrow \dfrac{ad-bc}{bd}\in \mathbb{Q}.[/imath] I meant that I emphasized the property and you emphasized the calculation. The result is the same.

 

[logistic_guy]

Not induction, inductive conclusion. It is the conclusion from the special case to the general case. This is allowed as long as we do not impose any restrictions on the special case. It is the opposite of a deductive conclusion where we conclude from the general case to the special case.

Deductive: The sum of a rational number and an irrational number is irrational, hence [imath] 1+\sqrt{2} [/imath] is irrational.

Inductive: Let [imath] r [/imath] be any rational number and [imath] i [/imath] any irrational number. Then ... [imath] r+i [/imath] is irrational and all sums of a rational number and an irrational number are irrational, hence [imath] 1+\sqrt{2} [/imath] is irrational.


It is the same as my proof, just with more words, an unnecessary restriction to integers, and a proposed second part for rationals. Maybe I should have said distract instead of hide. I called it closure under negation and addition, and you wrote [imath] \dfrac{a}{b}\, , \,\dfrac{c}{d}\in \mathbb{Q} \Longrightarrow \dfrac{ad-bc}{bd}\in \mathbb{Q}.[/imath] I meant that I emphasized the property and you emphasized the calculation. The result is the same.

I understood a lot from your explanation. Thanks professor. If our proofs are the same, I will just say that it's always good to solve the same problem with two different methods.

Why was my step an unnecessary restriction on integers? If I did not include that part, how would I show that \(\displaystyle \frac{ad - bc}{bd} = \frac{m}{n}\), where \(\displaystyle m, n \in \mathbb{Z}\)?

 

[fresh_42]

I would simply delete the sentence with the asterisk. It is not really necessary. [imath] \mathbb{Q} [/imath] is a field and therefore closed under addition and negation, and [imath] \mathbb{Q}-\{0\} [/imath] is closed under multiplication and inversion. It is not necessary to fall back on how the rational numbers are defined by integers. Otherwise, you could as well even go back to the natural numbers or even the Peano axioms. It is all a question of what can be expected as a common ground.

 

[logistic_guy]

Let me continue to solve \(\displaystyle (b), (c)\) and \(\displaystyle (d)\).

A counterexample is sufficient.

\(\displaystyle (b): \ \text{false}\), since \(\displaystyle \sqrt{2} + (-\sqrt{2}) = 0\), and \(\displaystyle 0\) is rational.

\(\displaystyle (c): \ \text{false}\), since \(\displaystyle 0 \times \sqrt{2} = 0\), and \(\displaystyle 0\) is rational.

\(\displaystyle (d): \ \text{false}\), since \(\displaystyle \sqrt{2} \times \sqrt{2} = 2\), and \(\displaystyle 2\) is rational.

 

[fresh_42]

Let me continue to solve \(\displaystyle (b), (c)\) and \(\displaystyle (d)\).

A counterexample is sufficient.

\(\displaystyle (b): \ \text{false}\), since \(\displaystyle \sqrt{2} + (-\sqrt{2}) = 0\), and \(\displaystyle 0\) is rational.

\(\displaystyle (c): \ \text{false}\), since \(\displaystyle 0 \times \sqrt{2} = 0\), and \(\displaystyle 0\) is rational.

\(\displaystyle (d): \ \text{false}\), since \(\displaystyle \sqrt{2} \times \sqrt{2} = 2\), and \(\displaystyle 2\) is rational.

Right. What would you get for (c) if you excluded zero as a rational number?

 

[logistic_guy]

Right. What would you get for (c) if you excluded zero as a rational number?

I will let \(\displaystyle x\) be a rational number and I will let \(\displaystyle y\) be an irrational number. I will assume \(\displaystyle xy\) is a rational number.

Then,

\(\displaystyle xy = \frac{a}{b}y = \frac{c}{d}\)

\(\displaystyle y = \frac{bc}{ad} \in \mathbb{Q}\)

This shows \(\displaystyle y\) as rational which is a contradiction. Therefore, \(\displaystyle xy\) is irrational.

If I use your method, I get:

Let [imath] r\in \mathbb{Q} [/imath] and [imath] i \in \mathbb{R}\setminus \mathbb{Q}. [/imath]

Assume [imath] r' := r \times i \in \mathbb{Q}. [/imath]

Then,

\(\displaystyle i = \frac{r'}{r} \in \mathbb{Q}\) which contradicts our assumption, so \(\displaystyle r \times i\) must be irrational.