Rational and irrational numbers. (semi- )

[Sven]

Rational and irrational numbers. (semi-urgent)

I need to figure this out by tomorrow =/

Homework Statement

a. If a is rational and b is irrational, is a+b necessarily irrational? What if a and b are both irrational?
b. If a is rational and b is irrational, is ab necessarily irrational?
c. Is there a number of a such that a^2 is irrational, but a^4 is rational?

Homework Equations

none.

The Attempt at a Solution

a. I think I have this first part. You can prove it by contradiction.

R= some rational number

a+b = R
b = R-a
A rational number minus a rational number is a rational number. This would mean b = rational, which is not true. therefore a+b is irrational.

This second part, if both are irrational? I was thinking:

a+b = R

a = R-b, or b = R-a. I'm not sure how this helps me x(
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b. If a is rational and b is irrational, is ab necessarily irrational?

No idea, but here's my attempt:

Two cases. If a = 0, ab = 0, and 0 is rational so ab is rational.

if a =/= 0...proof by contradiction maybe?

a*b = rational
a*b = a*b
b = a*b*a^-1?

But then b=b? And that doesn't help me.
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c Is there a number a such that a^2 is irrational, but a^4 is rational?

Well again I have no idea but here's my attempt:

a=b
a^2 = ab
ab = x
b = x/a
b = x * a^-1

...again no idea, please help. x(

 

[VietDao29]

a. I think I have this first part. You can prove it by contradiction.

R= some rational number

a+b = R
b = R-a
A rational number minus a rational number is a rational number. This would mean b = rational, which is not true. therefore a+b is irrational.

So far so good. :)

This second part, if both are irrational? I was thinking:

a+b = R

a = R-b, or b = R-a. I'm not sure how this helps me x(

When finding a proof seems hopeless, one should try to search for a counter-example.

According to part a,

$$1 + \sqrt{2}$$, and $$-\sqrt{2}$$ are both irrational, what if you take the sum of them?

b. If a is rational and b is irrational, is ab necessarily irrational?

No idea, but here's my attempt:

Two cases. If a = 0, ab = 0, and 0 is rational so ab is rational.

First case, okay. :)

if a =/= 0...proof by contradiction maybe?

a*b = rational
a*b = a*b
b = a*b*a^-1?

But then b=b? And that doesn't help me.

Second case:

Well, just do as you did in part a. Like this:

$$a \in \mathbb{Q} \backslash \{ 0 \} 0, b \notin \mathbb{Q}$$

Assume that
$$ab = r$$, where r is a rational number.
$$\Rightarrow b = ra ^ {-1}$$

What can you say about b in the above expression?

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Additional Problem:

If a, and b are both irrational numbers, is ab also irrational?

c Is there a number a such that a^2 is irrational, but a^4 is rational?

Well again I have no idea but here's my attempt:

a=b
a^2 = ab
ab = x
b = x/a
b = x * a^-1

...again no idea, please help. x(

Think about 4-th root. :)

 

[Architeuthis Dux]

Hello,

I am happy to help you understand the concept of rational and irrational numbers.

First, let's define these two types of numbers. Rational numbers are numbers that can be written as a ratio of two integers, while irrational numbers cannot be written as a ratio of two integers and have an infinite number of non-repeating decimals. Examples of rational numbers are 1/2, 0.75, and 3, while examples of irrational numbers are pi, e, and the square root of 2.

Now, let's address the first part of the question. If a is rational and b is irrational, is a+b necessarily irrational? The answer is yes. Your proof by contradiction is correct. Since a is rational and b is irrational, their sum must be irrational because adding a rational number to an irrational number will always result in an irrational number.

For the second part, if both a and b are irrational, is a+b necessarily irrational? The answer is still yes. Your approach of setting a + b = R is correct. Since a and b are both irrational, R (the sum of a and b) must also be irrational.

Moving on to the third part, is ab necessarily irrational if a is rational and b is irrational? The answer is no. Your approach of using proof by contradiction is correct. Let's say a = 2 (rational) and b = √2 (irrational). Their product, ab, is 2√2, which is irrational. However, if a and b are both irrational, then ab will always be irrational.

Lastly, is there a number a such that a^2 is irrational, but a^4 is rational? The answer is no. Any number raised to an even power will always result in a rational number. Let's take the example of a = √2. In this case, a^2 = (√2)^2 = 2, which is rational. And a^4 = (√2)^4 = 4, which is also rational. Therefore, there is no number a that satisfies this condition.

I hope this helps clarify the concept of rational and irrational numbers for you. Remember, rational numbers can always be written as a ratio of two integers, while irrational numbers cannot. Keep practicing and you will become more comfortable with these concepts. Good luck on your assignment!