# 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.

Note by Anik Mandal
3 years, 5 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 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"
MathAppears 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} $$\boxed{123}$$

Sort by:

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

- 3 years, 5 months ago

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

- 3 years, 5 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.

- 3 years, 5 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.

- 3 years, 5 months ago