Question

someone help me out. I am confuse on this math problem. How do i know if its true or false? I already tested out some values

Answer 1

3c

Answer 2

My work:

Answer 3

a divides (b-1)*(c-1)

Answer 4

it's working for me

Answer 5

let me draw

Answer 6

hmm did i make an error?

Answer 7

oh wait, i did a mistake on the last row. I just fixed it . Therfore this statement is true?

Answer 8

It's an AND statement right?

Answer 9

@xXMarcelieXx are you here?

Answer 10

yes

Answer 11

There's a way to know whether it is true or false before you even start creating your table

Answer 12

\((bc - 1)\) must be a factor of \((b - 1)(c - 1)\) which it is not.

Answer 13

So ultimately it will fail

Answer 14

You just have to find the value where it fails

Answer 15

you usually get the conclusion to test if it factors out?

Answer 16

I don't understand your question.

Answer 17

all i wanted to know was...when they say *(bc-1), do they mean (b-1)(c-1)..or is the expression is given as is...

Answer 18

no

Answer 19

meaning, when do we usually factor out things ?

Answer 20

We're not factoring anything. It's just that if \(a\) is dividing both \((b - 1)\) and \((c - 1)\) then surely it will divide the product of \((b - 1)(c - 1)\) but \((bc - 1)\) is not a factor of any of those numbers.

Answer 21

AAAAAAAAA icy now

Answer 22

i tried a = -2 both the individual and the product gave me the same answer....

Answer 23

so, is this statement false? I thought it was true

Answer 24

try more points

Answer 25

wait, let me draw.

Answer 26

Does it work for negative numbers?

Answer 27

@xXMarcelieXx always test all possibiities

Answer 28

wait, the values of *b* and *c* remain constant right ?

Answer 29

we're only changing the values of *a* here right??

Answer 30

Test a negative number for b and a positive number for c

Answer 31

see if that works

Answer 32

i tested negative numbers and it seems to be true but unless i did some algebra errors. like a= -3, b= -2 and c= -4

Answer 33

Test negative and positive numbers mixed

Answer 34

Don't test all positive and all negative

Answer 35

okay, i just tested some values like that and it turned out to be true for some reason

Answer 36

Yep, It's probably true. Just need to prove why

Answer 37

okay, i'll do it right now

Answer 38

I did a rough draft on it

Answer 39

my written proof

Answer 40

You forgot to define q

Answer 41

that's the only flaw I see

Answer 42

Actually I see another mistake

Answer 43

oh, hmm where do i define

Answer 44

Q*

Answer 45

Well the q seems to appear out of nowhere. It is not linked to the rest of your proof.

Answer 46

q is q= akr+k+r

Answer 47

Exactly. I'm pretty sure you need to define it explicitly somewhere.

Answer 48

But then I see that you have bc = aq and also that bc - 1 = aq

Answer 49

Both can't be true.

Answer 50

oh wait i see it. I just fixed it

Answer 51

Better but now you have an open parentheses left hanging

Answer 52

oh okay, just fixed it :)

Answer 53

The part about closed under multiplication is it in the right place? We're just defining q.

Answer 54

so should i delete it ? I think theres one where we say "since the integers are closed under addition or multiplication"?

Answer 55

Yes integers are closed under multiplication and addition, but there should be a reason for mentioning it.

Answer 56

It's not because you're defining q

Answer 57

oh i see. so then what can i say on for that one

Answer 58

Nothing more than what you've already said minus the integers closed under multiplication part

Answer 59

One second. Let me try something real quick

Answer 60

okie

Answer 61

Okay nevermind. You should be good to go. Let me know what feedback you get on that one.

Answer 62

Hang on, I did find a flaw

Answer 63

a can't be zero, but b or c can

Answer 64

Nope, nevermind, if b or c is 0, then a has to be 1 or -1

Answer 65

Goodnight

Answer 66

so do i just leave the statement " since the integers are closed under multiplication" sorry i got confused lol

Answer 67

It doesn't fit with the rest of the sentence.

Answer 68

It isn't used for defining variables

Answer 69

It's just used to state that if you multiply two integers the result will also be an integer.

Answer 70

Anyways, I'm off.

Answer 71

i am thinking it should be something like this then

Answer 72

no dont leave yet loll

Answer 73

I would write it as let q be an integer such that q = (akr + k + r). Since integers are closed under multiplication, we can say that aq is also an integer. Thus bc - 1 = aq. Therefore a|(bc - 1)

Answer 74

Something like that

Answer 75

oh okay, got it :)