Question

Suppose r, s are elements of the real numbers. Prove the following:
1. r is an element of the rational numbers and (r+s) is not rational implies s is not rational
2. r is not rational, s is not equal to 0, and rs is rational implies s is not rational

Answer 1

prove the contrapositive for #1 :
if s and r are rational, then r+s is rational

Answer 2

I thought the contrapositive would be claiming:
If s is rational \(\implies\) (r+s) is rational and r is not rational.
I thought the implication flipped and all the statements were negated.

Answer 3

I thought it was
If s is rational, then r is not rational or (r+s) is rational.

Answer 4

you know or instead of and

Answer 5

I guess it would make sense to know which is which first, haha

Answer 6

oops! yeah \(\large \neg(A\land B) = \neg A \lor \neg B\)

Answer 7

Okay, that makes more sense. But yeah, so if I list it out:
A = r is rational
B = r+s is not rational
C = s is not rational.

So I would want the contrapositive of (a^b) implies c, which is....
C v not(A^B)

Yeah, I forgot how to properly negate an implication, whoops x_x

Answer 8

I think you can prove the first one by contradiction.

Answer 9

i agree, contradiction seems natural to me in this case.

Answer 10

\[\text{ Assume } s=\frac{a}{b} , a,b \in \mathbb{Z} \text{ with } b \neq 0\]

Answer 11

Yeah, I wasn't sure which method worked. I think I misunderstood contrapositive, but either way I thought it looked bad. So, which part of the statement would I contradict? Is there a specific part of the statement that must be chosen for the contradiction, or is it up to me really? Like, yeah, I think the natural one would be to say s = a/b, but is that the choice that must be made for contradiction? Just a conceptual question.

Answer 12

If you want to prove \(A\Longrightarrow B\) by contradiction, you have to show that \(A\text{ and } \neg B\) leads to a contradiction. You don't get to choose what to negate. You have to negate the conclusion.

Answer 13

Even though there are 3 statements, we're counting the first two as being A?

Answer 14

Yep.

Answer 15

first two are premesis

Answer 16

So a proof by contradiction would start out as:
Assume \(r\) is rational, and \(r+s\) is not rational, and \(s\) is rational.

We play with this until we reach something that is false.

Answer 17

Okay, so we only negated the conclusion, gotcha. So do we still have to work in the direction of if -> then, or do we just want any sort of contradiction?

Answer 18

At this point, we want any contradiction. Using those three statements.

Answer 19

So if r is rational, it can be written as something like m/n and if s is rational, it can be written as something like p/q. So r + s is:\[\frac{ m }{ n }+ \frac{ p }{ q }\]But...what kind of expression do we have to represent irrational? Like, I can say rational is an expression m/n, but what is irrational?

Answer 20

m,q,n, and p are what kinda numbers?

Answer 21

And what happens when you combine those fractions?

Answer 22

Oh, integers where n and q are nonzero.

Answer 23

\[\frac{ m }{ n }+ \frac{ p}{ q }= \frac{ mq + np }{ nq }\]

Answer 24

is mq+np an integer?
is nq an integer?

Answer 25

Based on closure, I would think so...?

Answer 26

yeah but that would mean r+s is what kinda number?

Answer 27

Right, so r+s is an integer, but that's a contradiction. Alright, cool. So the only problem is this analysis class is REALLY picky about using only what we have been given. So I'm not sure whether or not I could make a closure argument. Is there another way it might be able to be stated?

Answer 28

r+s is a rational, but that's a contradiction*

Answer 29

Oops, lol.

Answer 30

looks we gave up cheaply on proving the contrapositive hmm

Answer 31

I wouldn't know which way is best, lol. But yeah, is there another way to come to the above result, that r+s is rational without making a closure argument? Because we haven't explicitly discussed closure.

Answer 32

i think contradiction and contrapositive take same effort

Answer 33

I didn't know how to do the contrapostive way

Answer 34

The contrapostive isn't a bad idea, it just sounds a little strange in my opinion.

Answer 35

oh you don't like saying
\(\large (n,q) \in \mathbb Q \implies nq \in \mathbb Q\)

Answer 36

If s is rational, then r is not rational or (r+s) is rational.


We get to suppose s is rational...But r isn't in the hypothesis so I don't know what to do with that. So I don't see a direct constrapositve proof

Answer 37

could be wrong

Answer 38

There is no way to do this proof without knowing the fact that the sum of two rational numbers is still a rational number. That is what we mean by closure. You don't have to specifically say "closure", but you do need that the sum of two rational numbers is a rational number.

Answer 39

^^

Answer 40

The issue that freckles has with the contrapositive is basically my qualm as well. It just sounds...strange haha.

Answer 41

It does work though. It would go like this.

Assume s is rational. We want to show that either r is irrational, or r+s is rational. If r is irrational, we are done. If not, then r is rational, and since s is rational, r + s is rational..

Answer 42

See how it just kind of sounds....weird? Its totally equivalent to a direct proof or a a proof by contradiction, but it just has a strange vibe to it O.o

Answer 43

nice :) the logical disjunction arguemnt... looks more of a proof in logic class !

Answer 44

That is neat.

Answer 45

Ah, okay, so that's how it would work with the contrapositive. Thanks very much, all of you, I really appreciate it ^_^

Answer 46

so if i were dirac, i would call contradiction proof an elegant one and the contrapositive proof a weirdo ;p

Answer 47

haha

Answer 48

I would agree :) Although one of my professors would yell at me for doing too many proofs by contradiction >.< He said that real men do things directly >.>

Answer 49

weirdos need to exist
where would we be without Einstein? i heard he was weird

Answer 50

lol i thought contrapositive is also a kind of indirect proof as we are switching around the statements and negating.. i think its more indirect than a simple contradiction.. i dont know what your prof would call a man doing all contrapositive proofs hmm

Answer 51

what if you want to do the a proof by contradiction of the contraspotive
would that be a indirectest proof

Answer 52

i'm being dumb
you don't have to answer that

Answer 53

haha, thats awesome.

Answer 54

thats a frecky idea haha! it should work atleast logically

a proof by contradiction of a contraapositive of a given statement will be a real turn on for a logic prof ;)

Answer 55

We would have if p then q.
Then turn it into not q then not p
then by contradiction we would assume p and not q

Answer 56

but if wanted to do if p then q by contradiction then we would assume not q and p
it would be logically equivalent for sure

Answer 57

Woah! that looks like a round trip, thats all i guess... its same as the contradiction of original statement's conclusion

Answer 58

right

Answer 59

it is a circle

Answer 60

that would be a question on a test..
i don't know how to phrase it exactly

Answer 61

good question*

Answer 62

Is it pointless to prove some thing by contradiction of the contrapositive?

Answer 63

That is the only way I can think of how to phrase it

Answer 64

lol im still thinking...

for p->q
both the statemetns in `direct contradiction` and `contradiction of contrapositive` are same : \(\large p \land \neg q\)

Answer 65

so `contradiction of contrapositive` may not ease out things that the original `direct contradiction` cannot... as the statement we start with in both proofs are same hmm

Answer 66

@freckles I apologize to call you back. I thought the discussion might give me an idea on the 2nd part of the problem, but it wasn't something that jumped to me. I figured it would be best to take the same approach, do negation and say s is rational. In the first part, we were able to combine fractions and make a rational expression. But it seems like the second problem would want us to construct an irrational expression. But I'm not sure what such an expression would look like. Any ideas?

Answer 67

try contradiction again

Answer 68

Yes, that is where I was going. The contradiction would be to assume s is rational though, correct? Since that is the conclusion. But I was looking at it and trying to see which of the statements I'd be able to somehow contradict. I'd have to somehow get either r to be rational, s = 0, or rs to be irrational. I don't know what an irrational expression would even look like. Yet I felt like somehow I would have to construct such a statement to get a contradiction.

Answer 69

\[s \text{ is rational } , rs \text{ is rational}, s \neq 0, \text{ and } r \text{ is not rational} \]

Answer 70

\[\text{ Let } a,b,c,d \in \mathbb{Z} \text{ with } b,d \neq 0 . \\ \text{ Let } s=\frac{a}{b} \text{ and } rs=\frac{c}{d}\]

Answer 71

\[rs=r \frac{a}{b}=\frac{c}{d}\]
Ans since s does not equal 0 then a does not equal 0
so \[r=\frac{c}{d} \frac{1}{s}=\frac{c}{d} \frac{b}{a}\]

Answer 72

what do you notice here?

Answer 73

Im not sure, lol. Like, I'm not sure if there is something funky about either of those expressions. I see the equality and how you got them, just not sure what is off about them immediately.

Answer 74

Oh, does that make R automatically rational?

Answer 75

r*

Answer 76

yes we have a product of integers on top and on bottom which would make r rational

Answer 77

Lol, thats actually kinda funny, I spent the whole time looking at things similar to that wondering how I was going to get something that looked irrational (to contradict rs is rational) when all I had to do is see that r was rational with those manipulations x_x

Answer 78

Sometimes I look at things forever too and wonder why they aren't making sense and then i give up and sometimes post and ganiesh always knows what to say.

Answer 79

or whatever his name is

Answer 80

Yeah. I mean, there's always logic to these that can be seen, I'm just not used to working with them and manipulating them yet.

And if I also may ask, when you and ganeshie were talking earlier, ganeshie mentioned logical disjunction argument...what is that by chance?

Answer 81

That is kinda a term I forgot and would have to look up

Answer 82

But he was referring to nerd's post

Answer 83

if i remember correctly

Answer 84

Yeah. I just hadn't heard of it. I've heard of disjoint, but not sure if thats the same.

Answer 85

Assume s is rational. We want to show that either r is irrational, or r+s is rational. If r is irrational, we are done. If not, then r is rational, and since s is rational, r + s is rational..

Answer 86

disjunction is a fancy name for "logical or"

Answer 87

lets mimic nerd's proof by contrapositive for part 2 also maybe ?

Answer 88

I wouldnt mind seeing how that works ^_^

Answer 89

statement :
r is not rational, s is not equal to 0, and rs is rational implies s is not rational

contrapositive :
s is rational implies "r is rational" or "s = 0" or "rs is not rational"

Answer 90

Yeah, had no idea how you'd state the contrapositive with 3 statements like that. Had no idea if you'd make them all "or" statements or not :P

Answer 91

if r is rational or s = 0, we are done. if not, rs = irrational*s = irrational \(\large \square \)

Answer 92

yes pack those 3 statements before implies as one :

` r is not rational, s is not equal to 0, and rs is rational` implies s is not rational

Answer 93

negation of AND of several several statements is OR of negation of each statemetn :

\[\large \neg(A\land B\land \cdots ) = \neg A \lor \neg B \lor \cdots \]

Answer 94

not(A and B)
=notA or notB
so if B=(C and D)
=notA or not(C and D)
=notA or (notC or notD)
=notA or notC or notD

Answer 95

^^

Answer 96

That contrapositive seems....off. Like, it doesnt seem like youre doing any sort of proof. I guess the 2nd part of the statement is using the idea that irrational * rational = irrational (I know it's true, but is that a some sort of actuak explicit property or is it a given?). But for the first part, yeah, if r is rational or s is 0 then it's true, but you don't have to really show anything for that?

Answer 97

we are translating the given problem into an equivalent statement

Answer 98

contrapositive is logically equivalent to given statement, so proving a contrapositive is same as proving the given statemetn.

so yes, the actuall proof starts only after taking the contrapositive