Question

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Answer 1

@ganeshie8

Answer 2

@terenzreignz

Answer 3

@freckles

Answer 4

Any ideas on where to start?

Answer 5

all I can grab is that we are restricted to a universe of reals (is this real numbers)
and then supposedly either there exists an A and exists a C such that for all B and for all D ad=bc

Answer 6

yes, it's real numbers, and yeah that's what it's saying

Answer 7

I'm getting rusty on this. Let me process...

Answer 8

Make it easier

Answer 9

When you have to prove that something exists, you have to produce it;
When you have to prove that something holds for all elements of a set, you have to use arbitrary elements from said set...

Answer 10

Actually, never mind.
Let a = c = 0
then ad = bc for all b and d in the reals.

Case closed.

Answer 11

This student is college.

Answer 12

How about it, @Angelina_Vella ?
Suspiciously simple?
I'm doubting myself lol

Answer 13

It feels right, haha it's what I was thinking before, but I can't see how it could be that short.

Answer 14

Wanna have some fun?
Let's assume none of them are zeros, maybe that will let you find some peace haha

Answer 15

Haha alright

Answer 16

Let b and d be nonzero real numbers.
Let c = 1.

Then take \[\Large a = \frac{b}{d}\]
This exists because d is nonzero.

Then \[\Large ad =\frac{b}{\cancel d}\cdot \cancel d = b = bc\]

And the assumption holds.

Answer 17

Ohhh I see it now

Answer 18

There was another question similar to this one,

Answer 19

This is complicated...

Answer 20

First I re-wrote it as products instead of fractions by cross multiplying, then are your supposed to set each side of the equation to one?

Answer 21

Not that simple. Best not to mess with fractions because you'd be implicitly making assumptions by cross-multiplying.

Namely, by cross multiplying, you are already assuming equality and that the denominators are nonzero

Answer 22

But the only way I see how you can proving that it is true is if both sides equal 1

Answer 23

and to prove it false they should be different or that they can never be equal

Answer 24

Let me think again...

Answer 25

This was your original question, wasn't it? :P
And you did the cross-multiplying and gave the question you initially posted?

Answer 26

No no, this is another question, sorry

Answer 27

the attached image is another question

Answer 28

I was just thinking the proof to this one would be similar to the one we did in the original posted question

Answer 29

well "explaination"

Answer 30

For proving the first part, yes, setting them to be equal to one seems sound.
For any \(\large a \in \mathbb N\), choose \(b = a\)

Answer 31

Look at \[ad = k\]

Can you find an integer \(d\) for all \(a\) ?
\(k\) is fixed here.

Answer 32

So that for any \(\large c \in \mathbb N\), you can choose \(d = c\)

Answer 33

I'm a little lost

Answer 34

Okay, let me rephrase what I said earlier:

When you have \(\forall \), then you have to taker arbitrary elements when you're proving.
When you have \(\exists\), then you can choose a specific element when you're proving.

Answer 35

Okay that makes sense

Answer 36

So, we move at them one at a time, starting from the left:
\[\large \forall a \ \ \exists b \ \ \forall c \ \ \exists d \ \ \frac{a}{b}=\frac cd \]

So...

Answer 37

\[\large \color{red}{\forall a} \ \ \exists b \ \ \forall c \ \ \exists d \ \ \frac{a}{b}=\frac cd\]

Let a be an arbitrary positive integer.

Answer 38

\[\large{\forall a} \ \ \color{red}{\exists b} \ \ \forall c \ \ \exists d \ \ \frac{a}{b}=\frac cd\]

Let a be an arbitrary positive integer.
Choose b to be equal to a.

Answer 39

Can you continue this trend, @Angelina_Vella ?
^_^

Answer 40

umm let c be an arbitrary positive integer that is....

Answer 41

equal to b?

Answer 42

No...
You're right about the c part.

Answer 43

oh okay and then let d be equal to c?

Answer 44

Yes. Exactly.
Then we demonstrate
\[\large{\forall a} \ \ {\exists b} \ \ \forall c \ \ \exists d \ \ \color{blue}{\frac{a}{b}=\frac cd}\]

Answer 45

Since a = b and c = d,\(\Large \frac a b = 1 = \frac c d\)
case closed. ^_^

Answer 46

Oh haha okay :) and then to prove the other false you have to show what a/b = c/d are never equal?

Answer 47

Sort of. But do you know how to negate quantifiers? ^_^

Answer 48

well they switch to the other dont they? kind of like laws of logic and deMorgan's law?

Answer 49

Exactly. The almighty negation. \(\Large \neg \)
We could try proving the negation of the statement \[\Large \forall a \ \ \exists d \ \ \forall c \ \ \exists b \ \ \ \ \frac a b= \frac c d\]

Answer 50

So here goes...
\[\Large \neg \forall a \ \ \exists d \ \ \forall c \ \ \exists b \ \ \ \ \frac a b= \frac c d\]
\[\Large \color{red}{ \exists a }\ \ \neg \exists d \ \ \forall c \ \ \exists b \ \ \ \ \frac a b= \frac c d\]\[\Large \color{red}{ \exists a }\ \ \color{red}{ \forall d} \ \ \neg \forall c \ \ \exists b \ \ \ \ \frac a b= \frac c d\]
\[\Large \color{red}{ \exists a \ \ \forall d \ \ \exists c} \ \ \neg \exists b \ \ \ \ \frac a b= \frac c d\]\[\Large \color{red}{ \exists a \ \ \forall d \ \ \exists c \ \ \forall b} \ \ \ \ \neg\frac a b= \frac c d\]\[\Large \color{red}{ \exists a \ \ \forall d \ \ \exists c \ \ \forall b \ \ \ \ \frac a b\ne \frac c d}\]

Answer 51

That's fancy

Answer 52

Only the best for you, lol
Now let me think some more...
Or you could do some thinking of your own XD

Answer 53

Haha well I never realized all you had to do was negate to make it not equal

Answer 54

But that was the easy part. Now comes the proving part.
First, we have \(\Large \exists a \)
We have to choose a very particular value for a.
Choose wisely... :D

Answer 55

Haha you kill me >.<

Answer 56

ummm

Answer 57

d? :o

Answer 58

No. See, we don't have a d, yet.
We're supposed to choose a value for a, FIRST, as this \(\exists a\) came before \(\forall d\)

Answer 59

oh okay, so like we choose a number?

Answer 60

Yeah... still thinking.
Just keep bouncing your ideas off me lol

Answer 61

so we need to have it so the values are different in the equation?

Answer 62

always different

Answer 63

so the main thing will probably be the last for every right?

Answer 64

brb in like ten minutes.
feel free to ask around lol

Answer 65

okie

Answer 66

So it's like an evil adversary problem, we have to pick variables associated with (there exists) and our evil adversary picks those associated with (for every)

Answer 67

@ganeshie8

Answer 68

Hmmmmmm

Answer 69

Yeah, I don't know how to finish this :/

Answer 70

Anyone understand what we're supposed to do to prove that a/b != c/d?

Answer 71

Okay... where was I? :D

Answer 72

need a particular value for (there exists) a

Answer 73

Ahh.
Pick a prime number.

Answer 74

like 2?

Answer 75

Prime numbers are convenient.
So let's choose a = 2

Answer 76

Now... let d be an arbitrary positive integer.
You guys catch me so far?

Answer 77

why do we choose a prime number?

Answer 78

Let me work out my "TJ Magic"

Answer 79

if d is an arbitrary positive integer then we're only considering the set \[Z^{+}\]

Answer 80

Yup. So uhh @Angelina_Vella
You following?

Answer 81

oh okay makes sense

Answer 82

for example
Z = { -3,-2,-1,0,1,2,3}
But Z plus = [0,1,2,3] those guys count

Answer 83

fffff... forgot the ... on each side XD

Answer 84

Zero is not included, Usuki-chan :P

Anyway, now we come to \(\exists c\)
Choose another prime number ^^

Answer 85

gomen XD

Answer 86

maybe we can keep it small like let c = 3?

Answer 87

Sure.
And finally, we have an arbitrary positive integer, b.

Answer 88

like b = 4 ?

Answer 89

Now, we have to prove that \[\Large \frac a b \ne \frac c d\]

Answer 90

No, Usuki-chan, we have \(\forall b\) so b has to be arbitrary

Answer 91

@Angelina_Vella
Still with me?

Answer 92

oh yeah that's for all of B

Answer 93

so far :)

Answer 94

so a = 2 and c = 3?

Answer 95

so what we have so far is
a = 2 c =3
and d = ? b = ?
for d and b are positive integers and a =2 and c = 3 are prime numbers