Proof Practice-I

$\text {Proof the following}$

$\bullet$ Given any positive integer $\displaystyle k$ ,prove that there are $\displaystyle k$ consecutive integers that are all composite.

#NumberTheory

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

One line proof. $(k+1)! + i$ where $2 \leq i \leq k+1$

Siddhartha Srivastava - 6 years, 3 months ago

Sorry,I didn't get it.Please explain it clearly@Siddhartha Srivastava

Anik Mandal - 6 years, 3 months ago

If u know factorial notation then notice that for each $i$ that same $i$ from factorial expansion will come out common and the expression will be composite.

Dinesh Chavan - 6 years, 3 months ago

Expanding on what Dinesh Chavan said. $(k+1)! = (k+1)*k*(k-1)...*4*3*2$ . So for each $2 \leq i \leq k+1$ , we see that $i|(k+1)!$ . Also, it is trivially true that $i|i$ . Therefore, $i| (k+1)! + i$ . Since $i$ divides the number and it is greater than one, the number must be composite.

Siddhartha Srivastava - 6 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$