Question

Prove by contradiction that the sum of irrational and rational is irrational

Answer 1

i suppose first step is to assume a and b to be the addends

Answer 2

then i do a = 2m/4n and b = x/y

Answer 3

then i assume the sum is rational

Answer 4

\[\frac{2m}{4n} + \frac xy \implies \frac{2m + 4x}{4n + 4y}\]

Answer 5

then...
\[\frac{2(m + 2x)}{4(n+y)}\]

Answer 6

this isn't simplified...so it's not rational

Answer 7

is that how you do it?

Answer 8

let \(i_1\) be an irrational number and \(r_1, r_2\) be rational numbers. Then, if you assume the inverse statement, we get:\[i_1+r_1=r_2\]This leads to:\[i_1=r_2-r_1\]But \(r_2-r_1\) is a rational number. Therefore...

Answer 9

i assume that means im wrong?

Answer 10

I couldn't follow your argument - where in your argument is the irrational number?

Answer 11

2m/4n

Answer 12

I'm confused about your assumptions. Why do you call a = \(\frac{2m}{4n}\)? Are m and n supposed to be integers? That would make a rational.

Answer 13

2m/4n does not represent an irrational number

Answer 14

isn't rational number a number that's not in simplified form?

Answer 15

No, a rational number cannot be expressed as a ratio of integers.

Answer 16

an irrational number is one that CANNOT be written as \(\displaystyle\frac{p}{q}\) where p and q are integers.

Answer 17

@SmoothMath Perhaps you meant 'can'.

Answer 18

but the proof of sqrt 2 is irrational involves using even/even

Answer 19

Thanks, Parth. A rational number can. An irrational number cannot.

Answer 20

I think you may be confusing tis with the proof that \(\sqrt{2}\) is irrational where we start by assuming it IS rational and in its SIMPLIFIED form.

Answer 21

yes i am confusing that

Answer 22

so 1/3 is a rational number. so is 12/24

Answer 23

2/4 is rational.

Answer 24

I see clarity in @asnaseer's proof. Cheers!

Answer 25

thanks @ParthKohli :)

Answer 26

i see short steps though...

Answer 27

?

Answer 28

Yeah, I agree with asnaseer. Now, for rigor, we could show that the difference of two rational numbers is rational. That's not hard to show though.

Answer 29

how do you rewrite irrational number then?

Answer 30

You cannot, lgba.

Answer 31

Yes @SmoothMath - that is why I ended with ... (to be completed by the asker of this question) :)

Answer 32

The reason I find asnaseer's proof easier is that it doesn't involve the \(\Large {p\over q}\) stuff.

Answer 33

To be a complete proof, it would have involved such kinds of things, although that part isn't too difficult.

Answer 34

part of me really thinks this is too short....

Answer 35

@lgbasallote - you just need to represent an irrational number by /some/ symbol. I used \(i_1\) to represent it.

Answer 36

@lgbasallote - do you want me to complete the proof for you?

Answer 37

rational - rational would be..

\[\frac ab - \frac xy\]
\[\implies \frac {ay - bx}{by}\]
so this is integer/integer

Answer 38

i can manage @asnaseer

Answer 39

Nailed it =)

Answer 40

yes - that is how to complete it. you have shown that \(r_2-r_1\) is rational.

Answer 41

The contradiction is that we get \(\rm i_1 = r_3\) where \(\rm r_3 = r_2 - r_1\) and irrational cannot be rational.

Answer 42

i envy the answerers of my questions though....they always get more than 3 medals....whilst i haven't gotten 3 medals in 1 question for a long time.....

Answer 43

@lgbasallote if you really want something to stick your teeth into, then try proving that:\[i_1^{i_2}\] can be rational. :)

Here \(i_1, i_2\) are irrational numbers.

Answer 44

i'm just in primary school...that is too advanced for me

Answer 45

Thanks LG. You stole my identity. :P

Answer 46

@lgbasallote if you really are in primary school then I declare you a genius! :)

Answer 47

why so?

Answer 48

Here in the UK, primary school means you are under 12 years old. And if you are doing problems like this at that age then you are indeed a genius!

Answer 49

hmm...time to migrate to UK....

Answer 50

:)