Rational Number Plus Irrational Number

mathxyz · 2024-05-05 08:36:24

Explain in your words why the sum of a rational number and irrational number must be irrational.

Thank you.

Bob · 2024-05-05 08:47:55

Assume the result is rational and rearrange to show an irrational is rational WHAT! That's known as proof by contradiction

Bob

mathxyz · 2024-05-05 14:10:05

Bob wrote:

Assume the result is rational and rearrange to show an irrational is rational WHAT! That's known as proof by contradiction
Bob

Can you show me what you mean using a simple example?

Bob · 2024-05-06 04:29:01

Not really because a proof needs to cover all cases so a number substitution won't do. I'll try to explain more carefully

If you add two fractions you get another fraction
The same is true for subtraction multiplication and division of two fractions
Mathematicians say the set of fractions ( or the rationals is the same thing) is CLOSED under the four operations meaning you don't get a different object.

What happens is you multiply a rational, say r, by an irrational, say i ?

Is the result a rational, an irrational or could it be either depending on what you start with. The question wants us to show it's an irrational.

There's a method of proof here that we can use called 'proof by contradiction". If there are two possibilities you assume the one you don't want and show this leads to an impossible result

So let's assume that r times i is another rational say s

ri = s

Divide by r and we get i = s/r

But by the rules for combining rationals this is another rational
So we end up with i = a rational
But it cannot be as we know it's irrational
So we have a contradiction and the assumption was false. So r times i isn't a rational so all that is left is to say it is irrational.

But hang on a mo. Let's check the proof carefully
At one stage I divide by r. Is that always OK
No, not if r is zero. So my proof doesn't work when r is zero

So what does happen then? Zero times i is zero which is rational

So the full conclusion is a rational times an irrational is always irrational unless the rational is zero

You can use the same type of proof to see what happens when the operation between the rational and irrational is add or subtract or divide

Bob

Phrzby Phil · 2024-05-06 06:43:25

I didn't know that i was considered an irrational.

Even if so, wouldn't the same "reductio ad absurdum" proof be better using a real irrational?

Last edited by Phrzby Phil (2024-05-06 06:45:29)

Bob · 2024-05-06 07:41:38

I just used i because irrational begins with an i.

Given my spell checker kept capitalising it I could have made a wiser choice. Ho hum.

Bob

Phrzby Phil · 2024-05-06 10:01:10

But your post has a lower case i, so as it stands it may be considered wrong, assuming the rational/irrational distinction applies to reals only.

I guess it's too late to correct, because then these comments will make no sense.

mathxyz · 2024-05-06 10:45:19

Bob wrote:

Not really because a proof needs to cover all cases so a number substitution won't do. I'll try to explain more carefully

If you add two fractions you get another fraction
The same is true for subtraction multiplication and division of two fractions
Mathematicians say the set of fractions ( or the rationals is the same thing) is CLOSED under the four operations meaning you don't get a different object.
What happens is you multiply a rational, say r, by an irrational, say i ?
Is the result a rational, an irrational or could it be either depending on what you start with. The question wants us to show it's an irrational.
There's a method of proof here that we can use called 'proof by contradiction". If there are two possibilities you assume the one you don't want and show this leads to an impossible result

So let's assume that r times i is another rational say s
ri = s
Divide by r and we get i = s/r
But by the rules for combining rationals this is another rational
So we end up with i = a rational
But it cannot be as we know it's irrational
So we have a contradiction and the assumption was false. So r times i isn't a rational so all that is left is to say it is irrational.
But hang on a mo. Let's check the proof carefully
At one stage I divide by r. Is that always OK
No, not if r is zero. So my proof doesn't work when r is zero

So what does happen then? Zero times i is zero which is rational
So the full conclusion is a rational times an irrational is always irrational unless the rational is zero

You can use the same type of proof to see what happens when the operation between the rational and irrational is add or subtract or divide

Bob

Thank you, Bob.

mathxyz · 2024-05-06 10:46:28

Phrzby Phil wrote:

I didn't know that i was considered an irrational.
Even if so, wouldn't the same "reductio ad absurdum" proof be better using a real irrational?

Why don't you show the rest of us your effort? We can then continue our discussion. Makes sense?

Bob · 2024-05-06 19:21:09

Don't worry. He's teasing me. I get it.

Bob

mathxyz · 2024-05-07 00:27:09

Bob wrote:

Don't worry. He's teasing me. I get it.
Bob

Teasing? Really?

Math Is Fun Forum

#1 2024-05-05 08:36:24

Rational Number Plus Irrational Number

#2 2024-05-05 08:47:55

Re: Rational Number Plus Irrational Number

#3 2024-05-05 14:10:05

Re: Rational Number Plus Irrational Number

#4 2024-05-06 04:29:01

Re: Rational Number Plus Irrational Number

#5 2024-05-06 06:43:25

Re: Rational Number Plus Irrational Number

#6 2024-05-06 07:41:38

Re: Rational Number Plus Irrational Number

#7 2024-05-06 10:01:10

Re: Rational Number Plus Irrational Number

#8 2024-05-06 10:45:19

Re: Rational Number Plus Irrational Number

#9 2024-05-06 10:46:28

Re: Rational Number Plus Irrational Number

#10 2024-05-06 19:21:09

Re: Rational Number Plus Irrational Number

#11 2024-05-07 00:27:09

Re: Rational Number Plus Irrational Number

Board footer