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proof by contradiction

not sure how to do this.Screenshot 2021-01-22 at 12.32.07.png

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Avatar for mqb2766
mqb2766
21
Original post by dnejfn
not sure how to do this.Screenshot 2021-01-22 at 12.32.07.png


Assume there is a greatest multiple, then show the contradiction.
It really would help to be specific about your problem.
Original post by mqb2766
Assume there is a greatest multiple, then show the contradiction.
It really would help to be specific about your problem.

i dont understand what they mean by greatest multiple
Original post by dnejfn
i dont understand what they mean by greatest multiple

N is a greatest multiple of 5 if
(a) N is a multiple of 5
(b) if M is a multiple of 5 it is always true that M<=N.
Original post by DFranklin
N is a greatest multiple of 5 if
(a) N is a multiple of 5
(b) if M is a multiple of 5 it is always true that M<=N.

if N is a multiple of 5 then let N= 5a
Reply 5
Avatar for mqb2766
mqb2766
21
Original post by dnejfn
if N is a multiple of 5 then let N= 5a

And the contradiction part ..
Original post by mqb2766
And the contradiction part ..

5a+1?
Original post by dnejfn
5a+1?

How is that a contradiction?
Original post by DFranklin
How is that a contradiction?

idk im so confused about this question
Original post by dnejfn
idk im so confused about this question


Look, numbers are infinite right? I'll give you an example that might help:
Prove by contradiction that there is no largest integer.

let us assume that there is a largest integer, N.
but when we add 1 to N we get N 1 and N 1>N. This contradicts our statement, so there is no largest integer.

These kinds of questions follow the same pattern, even if the context changes a little. Now how do you think we'll tackle your question?
(edited 3 years ago)
idk why the '+' doesn't show in my post.
I mean N + 1 and N + 1 > N
Original post by ilovephysmath
Look, numbers are infinite right? I'll give you an example that might help:
Prove by contradiction that there is no largest integer.

let us assume that there is a largest integer, N.
but when we add 1 to N we get N 1 and N 1>N. This contradicts our statement, so there is no largest integer.

These kinds of questions follow the same pattern, even if the context changes a little. Now how do you think we'll tackle your question?

so if we assume N is a multiple of 5, N = lets say 5a
but then when you add 1, 5a+1>N which contradicts the statement.
Original post by dnejfn
so if we assume N is a multiple of 5, N = lets say 5a
but then when you add 1, 5a+1>N which contradicts the statement.


but how would 5a + 1 be a multiple of 5? is 1 divisible by 5?
Original post by dnejfn
so if we assume N is a multiple of 5, N = lets say 5a
but then when you add 1, 5a+1>N which contradicts the statement.

But it's not a multiple of 5. You have to play by the rules of the game.
Original post by mqb2766
But it's not a multiple of 5. You have to play by the rules of the game.

so its 5a+5>N instead
Original post by dnejfn
so its 5a+5>N instead


Yes
Original post by dnejfn
so its 5a 5>N instead


Also, for future reference, keep it simple. Say something like:
N is the greatest multiple of 5. But since N 5 > N, and N 5 is also a multiple of 5, there is no greatest multiple of 5. Keep the variables to a minimum if you can help it. Avoiding clutter makes things clearer.
(edited 3 years ago)
Original post by ilovephysmath
Yes

so then that's the answer.
so if we assume N is a multiple of 5, let N= 5a

but then when you add 1, 5a+5>N which contradicts the statement. So there is no greatest multiple of 5.
Original post by dnejfn
so its 5a+5>N instead

It would be better to write
5(a+1) > 5a = N
Asumme N is the greatest multiple of 5

M = 5 x N = 5N

=> M>5N and M is also divisible by 5
Therefore, there is no greatest multiple of 5

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