Waste less time on Facebook — follow Brilliant.
×

A simple proof that 0=1

NEGATIVE 20=NEGATIVE 20
LETS BRAKE THESE NEGATIVE 20'S
16-36 AND 25-45
REWRITE LITTLE BIT MORE
4 SQUARE - 49=5 SQUARE - 59
LET US ADD A CONSTATNT BOTH THE SIDE
4 SQUARE - 49 +81/4 = 5 SQUARE - 59 +81/4
LET US FACTOR THIS NOW
4 SQUARE-249/2[9/2]SQUARE=5 SQUARE-259/2+[9/2]SQUARE
2 AND THE 9 GETS CANCELLED
FACTORISE MORE
[4-9/2] [4-9/2] = [5-9/2]] [5-9/2]
[4-9/2] SQUARE = [5-9/2] SQUARE
SAQARES AND 9/2 GET CANCELLED
SO 4=5
4-4 = 5-4 [ SUBTRACTING 4 FROM BOTH SIDE]
0=1

CAN YOU FIND ANY MISTAKE . PLEASE HELP ME IS THIS CORRECT?

Note by Sai Kiran
1 year, 8 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. 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 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"
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} \( \boxed{123} \)

Comments

Sort by:

Top Newest

49 and 59 are 49 and 59 respectively

Sai Kiran - 1 year, 8 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...