# largest

• May 29th 2009, 10:47 AM
Sampras
largest
Whats wrong with this proof showing that \$\displaystyle 1 \$ is the largest natural number?

Let \$\displaystyle n \$ be the largest natural number and suppose \$\displaystyle n \neq 1 \$. Then \$\displaystyle n >1 \$ so that \$\displaystyle n^2 > n \$; thus \$\displaystyle n \$ is not the largest natural number.

Basically the premises are false because there is no largest natural number? So you can derive a true conclusion from false premises? In other words, the argument is not sound?
• May 29th 2009, 11:25 AM
Moo
Quote:

Originally Posted by Sampras
Basically the premises are false because there is no largest natural number?

Because in a proof by contradiction, you have to assume that the property is true, which is not what you did, since you said that \$\displaystyle n\neq 1\$. And you have to conclude with a contradiction.

Quote:

So you can derive a true conclusion from false premises?
See the truth table for \$\displaystyle A\Rightarrow B\$
If A is false, then B can be either false or true, the proposition "\$\displaystyle A\Rightarrow B\$" will always be true.

In order to prove the property, you can just use the property of addition of natural numbers :
If a and b are natural numbers, then a+b is also a natural number.
Hence 1+1 is a natural number. And 1+1>1. Thus the statement is false (Surprised)
• May 29th 2009, 12:31 PM
Sampras
Quote:

Originally Posted by Moo
Because in a proof by contradiction, you have to assume that the property is true, which is not what you did, since you said that \$\displaystyle n\neq 1\$. And you have to conclude with a contradiction.

See the truth table for \$\displaystyle A\Rightarrow B\$
If A is false, then B can be either false or true, the proposition "\$\displaystyle A\Rightarrow B\$" will always be true.

In order to prove the property, you can just use the property of addition of natural numbers :
If a and b are natural numbers, then a+b is also a natural number.
Hence 1+1 is a natural number. And 1+1>1. Thus the statement is false (Surprised)

No I wanted to prove that \$\displaystyle 1 \$ was the largest number. So I assumed for contradiction that it was not. It all boils down to that the premises were false to begin with.
• May 29th 2009, 05:50 PM
Sampras
Its not the proof by contradiction that is wrong. It's the premises which are false which makes this proof bad. Is this correct?
• May 30th 2009, 07:03 AM
HallsofIvy
What "premises" are you talking about?

You prove directly that no number other than 1 can be the "largest natural number". That 1 is the "largest natural number" follows from that only if you are using the "hidden" assumption that there is a "largest natural number. That is false. Was that the premise?
• May 30th 2009, 07:37 AM
Sampras
Quote:

Originally Posted by HallsofIvy
What "premises" are you talking about?

You prove directly that no number other than 1 can be the "largest natural number". That 1 is the "largest natural number" follows from that only if you are using the "hidden" assumption that there is a "largest natural number. That is false. Was that the premise?

Yes that was the premise. And that premise was false. The point was that the proof by contradiction was set up correctly.
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