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I was shown the following proof by contradiction, which somehow shows that $47$ is the largest whole number. Obviously this is not true, but I am not exactly sure where the proof goes wrong. Here it is:

Assume $47$ is not the largest whole number. Let $n$ be the largest whole number. $$n > 47 \implies n-47>0 \implies (n-47)^2 > 0 \implies n^2 - 94n + 2209 > 0 \\ \implies n^2-93n+2209>n$$

Since we defined $n$ to be the largest whole number, the inequality $n^2-93n+2209>n$ is false. Therefore we have reached a contradiction thus proving $47$ to be the largest number.

My thoughts: It seems like this "proof" can be done with any number (ie $47$ does not serve a special purpose in the proof). This leads me to believe that defining $n$ be the largest number is the mistake. Is my evaluation correct?

Madhav Nakar
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  • Correct. It shows there is no largest number. – max_zorn Jan 26 '18 at 23:07
  • You assume on contrary that $;n;$ is the largest integer...but you could as well assume that elephants fly and are pink : there is no such thing as the largest integer number. – DonAntonio Jan 26 '18 at 23:07
  • It works with all numbers, and some numbers happen to be larger than others, and you can prove this with other proofs. – Yash Jain Jan 26 '18 at 23:09
  • within the past few weeks someone asked about existence of solutions in the calculus of variations. Until Weierstrass proved there was a difficulty, mathematicians routinely assumed that any reasonable looking PDE had a solution. Your "proof" is a very clever illustration of the nonsensical outcome. – Will Jagy Jan 26 '18 at 23:19
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    If $n$ is the largest whole number, then $n+1>n$ which is a contradiction. – Somos Jan 27 '18 at 01:40

2 Answers2

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"Let $n$ be the largest whole number ..."

"$47$ is not the largest number" isn't the only assumption you're making in the proof. When you bring $n$ into the picture, you're implicitly assuming that such an $n$ exists in the first place - that is, that there is a largest whole number.

Noah Schweber
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    Suppose we assume Donald Trump, rather than $47,$ is the largest whole number. We have a theorem saying whole numbers do not live in the White House. Thus we get a contradiction. – Michael Hardy Jan 26 '18 at 23:27
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The assumption " Let n be the largest whole number" implies the existence of a largest whole number which is proved to be false. From a false predicate you can come to any conclusion whatsoever.

All you are proving is that " If there exists a largest whole number then that largest whole number is 47 or any other number". Therefore you implicitly proved that there is no largest whole number.

Mohammad Riazi-Kermani
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