But the doesn"t really occupational does it? What"s a very basic proof (no number concept or any kind of sort prefer that)?


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Your proof is good. However, I check out some areas where you can add an ext detail:

Why space we assuming the $x$ is the largest herbal number? (Answer: because we room doing a proof by contradiction.)Why is the fact that $x + 1 > x$ relevant? (Answer: because it contradicts the fact that $x$ is the largest natural number.)

If we add these details in, the evidence looks like this:

We will display that there is no largest herbal number. In bespeak to perform this, let"s assume that $x$ is the largest natural number. Through the definition of "largest", over there is no herbal number i beg your pardon is higher than $x$. However, $x + 1$ is a natural number and $x + 1 > x$. This is a contradiction. QED.

You are watching: What is the largest natural number

There"s one an ext question that this proof doesn"t answer:

How carry out we know that $x + 1 > x$?

If friend haven"t currently proved the $x + 1 > x$, climate you must think around giving the a shot.


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answer Dec 12 "16 in ~ 23:00
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Tanner SwettTanner Swett
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I to be a viewpoint major, not dearteassociazione.orgematics, all the same, the complying with will prove what you space looking for:

Assume the visibility of a largest natural number (p)Being the largest natural number, "p" cannot be followed by a higher natural number. Meaning:

p + 1 ≤ p

Now subtract "p" from both sides. The result is:

1 ≤ 0

This has displayed that the presumption of "p" leader to a contradiction.

Therefore, that is not the situation that a largest natural number exists.

QED


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answer Nov 7 "17 at 14:10
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TaylorTaylor
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Here is a constructive proof.

Let $P(n)$ it is in the residential property that over there exists a herbal number $n"$ wherein $n" > n$. We prove this by induction.

Base case - $n=1$: Clearly, $P(1)$, due to the fact that $484000 > 1$.

Step – assume for $n$, prove because that $n+1$: assume by induction theory that over there exists an $n""$ such the $n"" > n$. By the precongruence properties of $>$ we quickly see that $n""+484000 > n+1$. So take $n" = n"" + 484000$.

If we desire to prove the statement that it is not the instance that over there exists one $n_0 \in \dearteassociazione.orgbbN$ such the $n \leq n_0$ for every $n \in \dearteassociazione.orgbbN$, then us assume that there exists such an $n_0$. However then we have actually $n_0 + 484000 \leq n_0$, i m sorry is a contradiction. Therefore $n_0$ can not exist.


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edited Dec 12 "16 at 22:45
answered Dec 12 "16 at 22:42
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Hans HüttelHans Hüttel
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Maybe that can be displayed in a million the ways, but here is one proof just for fun (which surely lacks some foundational rigor). We will usage this facts:

1) $1$ is not the largest herbal number

2) because that every organic number $n \neq 1$ the number $n^2-n+1$ is a herbal number the is no equal to $n$ (that is, we have $n^2-n+1 \neq n$)

3) $(n-1)^2>0$ for every organic $n \neq 1$.

Suppose there exists largest natural number, contact it $n_L$, now kind a natural number $n_L^2-n_L+1$. Since $n_L$ is the largest and also because the 2) we have: $n_L>n_L^2-n_L+1$, which is equivalent to $0>(n_L-1)^2$ however this is not possible because $n_L \neq 1$ and because that 3).


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reply Dec 12 "16 at 23:09
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