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#1
not sure how to do this.
0
4 months ago
#2
(Original post by dnejfn)
not sure how to do this.
Assume there is a greatest multiple, then show the contradiction.
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#3
(Original post by mqb2766)
Assume there is a greatest multiple, then show the contradiction.
i dont understand what they mean by greatest multiple
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4 months ago
#4
(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.
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#5
(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
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4 months ago
#6
(Original post by dnejfn)
if N is a multiple of 5 then let N= 5a
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#7
(Original post by mqb2766)
5a+1?
0
4 months ago
#8
(Original post by dnejfn)
5a+1?
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#9
(Original post by DFranklin)
0
4 months ago
#10
(Original post by dnejfn)
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?
Last edited by Aloe Striata; 4 months ago
0
4 months ago
#11
idk why the '+' doesn't show in my post.
I mean N + 1 and N + 1 > N
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#12
(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
0
4 months ago
#13
(Original post by dnejfn)
so if we assume N is a multiple of 5, N = lets say 5a
but how would 5a + 1 be a multiple of 5? is 1 divisible by 5?
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4 months ago
#14
(Original post by dnejfn)
so if we assume N is a multiple of 5, N = lets say 5a
But it's not a multiple of 5. You have to play by the rules of the game.
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#15
(Original post by mqb2766)
But it's not a multiple of 5. You have to play by the rules of the game.
0
4 months ago
#16
(Original post by dnejfn)
Yes
0
4 months ago
#17
(Original post by dnejfn)
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.
Last edited by Aloe Striata; 4 months ago
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#18
(Original post by ilovephysmath)
Yes
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.
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4 months ago
#19
(Original post by dnejfn)
It would be better to write
5(a+1) > 5a = N
0
4 months ago
#20
- prove by contradiction that there is no greatest multiple of 5

N = greatest multiple of 5

10 = multiple of 5 (2 * 5 = 10)

N > 10

10 * N cannot be greatest multiple of 5 than N because N is the greatest multiple 5. N > 10N but 10 * N is greater than N... so we have a contradiction.
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