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# The Proof is Trivial! watch

1. (Original post by rayquaza17)
I would say 0 is in the set of non-negative integers.

But Wolfram Alpha disagrees with what they define as the non-negative integers: http://mathworld.wolfram.com/NonnegativeInteger.html
0 is a non-negative integer. And Wolfram Alpha agrees with that...
2. (Original post by demigawdz)
im so aweful at maths i cannot even do this one.
lemme guess its probably an easy one
Well, I found it pretty easy tbh, but don't think you're awful because you can't do it. How much BMO have you done before?

Hint:
Spoiler:
Show
Think about what happens when n=m=0, and then use the first condition
3. (Original post by Renzhi10122)
In the question, I guess its the set of non-negative integers then.
It probably doesn't matter which definition you take, although I can do the question if you want to know.
4. (Original post by TheMagicMan)
It probably doesn't matter which definition you take, although I can do the question if you want to know.
If it doesn't take too long then thanks. Is there only one answer (just want to check)? Also, I wasn't the one who needed help with this one.
5. (Original post by Renzhi10122)
Well, I found it pretty easy tbh, but don't think you're awful because you can't do it. How much BMO have you done before?

Hint:
Spoiler:
Show
Think about what happens when n=m=0, and then use the first condition
quite a few questions but i was stumped by the 0 thingy.
if i considered it for example where it shouldngt be then i was gonna be horribly wrong from the get go (i worried)
But apparantly not from your hint.

BTw- I worked out when n=0,1 m=0,1 and when n=m=0,1
6. (Original post by TheMagicMan)
0 is a non-negative integer. And Wolfram Alpha agrees with that...
Yeah but the BMO thing was saying Z+ is the non-negative integers when that's the positive integers

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7. (Original post by rayquaza17)
Yeah but the BMO thing was saying Z+ is the non-negative integers when that's the positive integers

Posted from TSR Mobile
yeah I see the confusion now. i would say that as the words used are 'the non-negative integers' you should include zero, especially considering the obvious way to solve the problems involves using zero, although there are certainly other methods that don't need it.
8. (Original post by TheMagicMan)
yeah I see the confusion now. i would say that as the words used are 'the non-negative integers' you should include zero, especially considering the obvious way to solve the problems involves using zero, although there are certainly other methods that don't need it.
plz share these other methods
9. (Original post by demigawdz)
quite a few questions but i was stumped by the 0 thingy.
if i considered it for example where it shouldngt be then i was gonna be horribly wrong from the get go (i worried)
But apparantly not from your hint.

BTw- I worked out when n=0,1 m=0,1 and when n=m=0,1
Sorry, I don't understand the last line?
10. (Original post by demigawdz)
plz share these other methods
For example

Spoiler:
Show

Restate the first condition as

Then inductively

and it's pretty much done using the second condition.

OR

Spoiler:
Show

you could also use the second condition to say that f is strictly increasing (and therefore injective) and then apply f to the second condition and again it is pretty much done

11. (Original post by Renzhi10122)
Sorry, I don't understand the last line?
i put the values when n=0 and when n=1 into the functional eqns.
likewise for m.
likewise for when n=m (for when either ='s 1 or o)
In retrospect all those particular solutions seemingly made it hard to see what was going on lol
12. (Original post by demigawdz)
i put the values when n=0 and when n=1 into the functional eqns.
likewise for m.
likewise for when n=m (for when either ='s 1 or o)
In retrospect all those particular solutions seemingly made it hard to see what was going on lol
From doing n=m=0, you should be able to just compete the question. You should get
Spoiler:
Show
and then use the first condition
13. (Original post by demigawdz)
...
I read this book a while ago, and found it pretty decent.
14. (Original post by Lord of the Flies)
I read this book a while ago, and found it pretty decent.
thanks, does it have quite a lot of maths in it though????
the preview only shows history of functional eqns lol
15. (Original post by demigawdz)
Thanks, does it have quite a lot of maths in it though?
16. (Original post by Lord of the Flies)
Is it similar to the BMO handbooks?
17. (Original post by demigawdz)
Is it similar to the BMO handbooks?
No clue. Naturally, it contains quite a few IMO/Putnam problems though.
18. (Original post by Lord of the Flies)
No clue. Naturally, it contains quite a few IMO/Putnam problems though.
but is it worth forking out like £33 for it though?
19. Groups A and B have the same order and have the same number of subgroups, are they necessarily isomorphic? If this is true, can I use it in an A-Level MEI FP3 answer without explaining it?
I came across a past paper question asking me to explain why all groups of order 5 are always isomorphic to eachother. The first thing I wrote down was "5 is a prime number so by Lagrange's Theorem, the only subgroups it can have is itself and a group only containing its identity"

So the second part of my answer is "So all order 5 groups have the same number of subgroups, therefore all of them are isomorphic to eachother". But I am not sure if this is correct.
20. (Original post by Primus2x)
Groups A and B have the same order and have the same number of subgroups, are they necessarily isomorphic? If this is true, can I use it in an A-Level MEI FP3 answer without explaining it?
I came across a past paper question asking me to explain why all groups of order 5 are always isomorphic to eachother. The first thing I wrote down was "5 is a prime number so by Lagrange's Theorem, the only subgroups it can have is itself and a group only containing its identity"

So the second part of my answer is "So all order 5 groups have the same number of subgroups, therefore all of them are isomorphic to eachother". But I am not sure if this is correct.
There are groups satisfying that property but to my knowledge there are no elementary counterexamples.

Some research has led me to the fact that the modular group order 16 and is a counterexample. (http://www.opensourcemath.org/gap/small_groups.html)

You should never have to resort to such a theorem in A-level though.

One final thought that occurred to me about order 16 groups. There are 14 non-isomorphic groups order 16, and each of these groups can have at most 10 subgroups (itself, the 5 groups order 8, the 2 groups order 4, the 1 group order 2 and the one group order 1) so of course order 16 will contain a counterexample to your proposition! Indeed there are 51 groups order 32 so there will be an order 32 counterexample too. And so on.

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Updated: February 22, 2018
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