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this holds true for all integers such that: (a/a-1)+a=(a/a-1)*a, for all integer values of a. This was what I found while solving maths today morning.

Note by Siddharth Singh 3 years, 4 months ago

Easy Math Editor

*italics*

_italics_

**bold**

__bold__

- bulleted- list

1. numbered2. list

paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

> 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"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

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Hi

Try using \(\LaTeX\) !

" \dfrac{a}{a-1} + a = \dfrac{a}{a-1} \cdot a " produces \[ \dfrac{a}{a-1} + a = \dfrac{a}{a-1} \cdot a\]

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewestHi

Try using \(\LaTeX\) !

" \dfrac{a}{a-1} + a = \dfrac{a}{a-1} \cdot a " produces \[ \dfrac{a}{a-1} + a = \dfrac{a}{a-1} \cdot a\]

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