Divide rs 2379 into 3 parts so that their amount after 2,3 and 4 years respectively may be equal, the rate of interest being 5% per annum at simple Interest, what will be first Part
I have Tried
amount will be equal for all the three parts for n=2,n=3 and n=4 Amount = Simple Interest + Principal 2379/3= will give the value for each part it constitutes 793 , we cannot directly divide that three parts May be another three different types of Number
let the first part Number part i will be take x, surely the second part will be 2379-x and how will take the value for third part,Please anyone share the Answer and Logi
@user36188: You had 10 upvotes and 2 downvotes. You included your attempt in your original question, which is good. But you didn't use proper sentences and punctuation and didn't check your spelling. Furthermore, you didn't use LaTeX, as instructed by the FAQ which you were told to read before posting. Thus the downvotes were justified.
@MartinSleziak: I would like to but since I've interacted with this crankpot quite a lot in trying to encourage them to actually start learning logic, I can't vote to close and delete without it seeming like I'm enforcing my opinion.
@MartinSleziak: The questions themselves are nearly gibberish, but I can guess the underlying ideas, which laymen are bound to have without a proper logic foundation. I guess I'll answer to that idea one more time and see how it goes.
Take any non-empty set S <= N. If not exists m in N ( forall x in S ( m <= x ) ): | Let P(k) assert that forall x in S ( k <= x ). | Then P(0). | Given any k in N such that P(k): | | ... | | P(k+1). | Therefore P(k) for every k in N. | Let c in S. | Then P(c+1). | Thus c+1 <= c. | Contradiction. Therefore exists m in N ( forall x in S ( m <= x ) ).
I'll leave you to figure out the missing part. It's where you need to use the "no minimum".
That proves existence of a minimum. In this case it is easy to prove uniqueness from that.
P(0) is trivially true, and says nothing about minim elements.
*minimum
Take any non-empty set S <= N. If not exists m in N ( forall x in S ( m <= x ) ): | For each k in N let P(k) assert that forall x in S ( k <= x ). | Then P(0). | Given any k in N such that P(k): | | ... | | P(k+1). | Therefore P(k) for every k in N. | Let c in S. | Then P(c+1). | Thus c+1 <= c. | Contradiction. Therefore exists m in N ( forall x in S ( m <= x ) ).
Oops.
Sorry typo.. Corrected version below.
Take any non-empty set S <= N. If not exists m in S ( forall x in S ( m <= x ) ): | For each k in N let P(k) assert that forall x in S ( k <= x ). | Then P(0). | Given any k in N such that P(k): | | ... | | P(k+1). | Therefore P(k) for every k in N. | Let c in S. | Then P(c+1). | Thus c+1 <= c. | Contradiction. Therefore exists m in S ( forall x in S ( m <= x ) ).
What I said about P is correct. I just accidentally put "N" instead of "S" in the assumption.
I purposely didn't want to say "replace" or "substitute" because although that's the programmer's way (systematic and excellent), you must first grasp the meaning.
If you haven't read my post about Fitch-style natural deduction for quantifiers, you should. It gives you the systematic rules for deduction using them, including this one.
I always try to balance your reasoning out. Sometimes you think syntactically and miss the semantics (meaning). Sometimes you want an intuitive visualization but it may be easier to just chase the symbols.
There is a visualization though.. Just follow the induction.
At first we know that 0 is a lower bound.
So 0 can't be in S, otherwise it would be a minimum of S.
So 1 is a lower bound.
So 1 can't be in S, otherwise it would be a minimum of S.
Nobody does it the way I do, although there are similar variants. Unfortunately there is no online reference I could find.
Just look at the rules I gave; you're a programmer so it should be no trouble for you.
Notice that there is only one way to derive a sentence of the form "forall x in S ( P(x) )".. You must open a new subcontext "Given x in S".. and the proof will always look like:
Given x in S: | ... | P(x). forall x in S ( P(x) ).
I want you to attempt to fill in the gap in the induction proof here in the format of natural deduction.
Let me reproduce the outline with a bit more detail so that you can see where this is going.
If not exists m in S ( forall x in S ( m <= x ) ): | For each k in N let P(k) assert that forall x in S ( k <= x ). | Then P(0). | Given any k in N such that P(k): | | forall x in S ( k <= x ). | | If k in S: | | | ... | | | Contradiction. | | k notin S. | | Given x in S: | | | k <= x. | | | k != x. | | | k+1 <= x. | | forall x in S ( k+1 <= x ). | | P(k+1). | Therefore P(k) for every k in N. | Let c in S. | Then P(c+1). | Thus c+1 <= c. | Contradiction. Therefore exists m in S ( forall x in S ( m <= x ) ).
The third step is to let (hk)k∈N(hk)k∈N be a sequence of functions such that hkhk witnesses P(k)P(k) for every k∈Nk∈N, and show that hihi and hjhj agree on {0..i}{0..i} for every i,j∈Ni,j∈N such that i<ji<j. Again, this is by induction ('on jj').
Step 3 first does for each k in N let h[k] : {0..k}->E such that ( h(0) = c and forall n in N[<k] ( h(n+1) = g(h(n)) ) ) ) Note that this is valid because we proved uniqueness. Here I'm reusing the variable name "h" and this time it is a sequence, which is nothing more than a function with domain N.
Although this is a chat-room, you should save whatever you wish to keep, because it's a waste of time for me to keep searching for what I've typed previously. As you can see, it's troublesome to find it. I want the later version:
Given set E and function g : E->E and c from E: | [Steps 1,2,3,4] Step 1 proves forall k in N ( exists h : {0..k}->E ( h(0) = c and forall n in N[<k] ( h(n+1) = g(h(n)) ) ) ) Step 2 proves forall k in N ( forall h,h' : {0..k}->E ( h(0) = c and forall n in N[<k] ( h(n+1) = g(h(n)) ) ) and ( h'(0) = c and forall n in N[<k] ( h'(n+1) = g(h'(n)) ) ) implies h = h' ) Step 3 first does foreach k in N let H[k] : {0..k}->E such that ( H[k](0) = c and forall n in N[<k] ( H[k](n+1) = g(H[k](n)) ) ) ) by lots of set-theoretic machinery and then proves
Ok that allows us to do the first part of Step 3 which is to construct a sequence of functions H[k] for k in N.
The second part is needed for Step 4 to work.
Whether or not you classify it under Step 3 is a matter of taste.
in Step 4 we took the union of the whole sequence of functions. It is not obvious that the result is a function!
That is what the second part of step 3 is for. In general when we have a sequence of functions where the appropriate restrictions agree, the union will also be a function.
In general, it goes: (1) Build a sequence of approximations; (2) Prove that any term in the sequence contains every preceding term; (3) Take union to get the result.
It applies not just to functions.
Note that for functions f,g to agree on Dom(f), it means that f is a subset of g. This is how it fits the general technique.
Point out the step that fails. Each step is justified by the ZF axioms. If you disagree with ZF, then you'll have to pick another formal system. The first part of my answer at math.stackexchange.com/a/1893865/21820 is what you might expect to see in a system that does not have the notion of infinite sets.
If you already accept the notion of N as a single collection of all natural numbers, then you essentially have accepted enough of ZF to be satisfied by the ZF proof.
(1) Build a sequence of approximations; (meaning that you construct the approximations according to ZF, so, they're valid) (2) Prove that any term in the sequence contains every preceding term (proved by recurrence, so it was proved by a valid theorem, so true in ZF) (3) Take union to get the result (since the union is an axiom of ZF it's valid to do it).
But if you don't accept that there is a single entity that is infinite, then you can't take the union so simply, and I think you need to have a notion of procedures that may not halt, which is the way my first proof goes.
Notice that in that proof we construct a procedure (treat it as the description itself), which we can perform for any finite number of steps.
