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Prove by contradiction.

If sum of $N$ natural numbers is less than $N+2$ then each of these numbers is less than $3$.

Attempt:

Assume if the sum of $N$ natural numbers is less than $N+2$ then each of these numbers is greater than $3$.

Let $x$ be the sum of $N$ natural numbers

$x < N + 2 $

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    $\begingroup$ No, the negation of "all the numbers are less than $3$" is not "all the numbers are greater than $3$". $\endgroup$ – Andrés E. Caicedo Dec 8 '13 at 0:49
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Try this:

Assume the sum of $N$ natural numbers is less than $N+2 \implies$ at least one of these numbers is not less than $3$.

Call that number $Y$. But then $Y+2 \ge 3+2=5.$ But $5 > N+2 \implies$ Contradiction!

$\therefore$ the sum of $N$ natural numbers is less than $N+2 \implies$ each of these numbers is less than $3$.

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You've got the negation incorrect: There would only have to be one that's greater than $3$ to make the original statement false. As a hint towards proving the statement, though, try:

Suppose that the first one was at least $3$. Then the sum of all of them is at least

$$3 + \underbrace{1 + \dots + 1}_{n - 1 \text{ times}}$$

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The statement that you’re trying to prove is that if $x_1+x_2+\ldots+x_N<N+2$, where each $x_k$ is a natural number, then $x_k<3$ for $k=1,\ldots,N$. You appear to be trying to prove it by contradiction, which is fine, but the negation of

each of the numbers is less than $3$

is not

each of the numbers is greater than $3$.

It is not even

each of the numbers is greater than or equal to $3$.

The negation of

each of the numbers is less than $3$

is

at least one of the numbers is greater than or equal to $3$.

Assume that at least one of the numbers is $\ge 3$; the other $N-1$ numbers are at least $1$, so the total must be at least ... what?

(Note that this problem requires you to assume that natural number means positive integer. For many of us that’s false, because we include $0$ as one of the natural numbers.)

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