Here are some examples of false Mathematical proofs:

Number 1:

Number 2:

Number 3 (the famous one):

let:

Number 4:

(1-1) + (1-1) + … + (1-1) + (1-1) + … = 0

1 + (-1+1) + (-1+1) + … + (-1+1) + … = 0

1 + 0 + 0 +… + 0 + .. = 0

1 = 0

Prove that:

Solution:

First,

It follows that:

Then,

Hence,

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5 040 – Greek philosopher Plato gave this number as the population of an ideal city.

This number has several factors, which implies an efficient division of population for the sake of land distribution, taxes, war etc.

In fact,

5, 040 = 7! = (1)(2)(3)(4)(5)(6)(7)

5, 040 = (2⁴)(3²)(5¹)(7¹)

Thus, 5 040 has (4+1)(2+1)(1+1)(1+1) = 60 factors

Compute in (less than) 30 seconds without using calculator.

My technique:

Since 2007 is the arithmetic mean of the numbers 2004, 2006, 2008 and 2010, then: Let x = 2007

So, x-3 = 2004, x-1 = 2006, x+1 = 2008, x+3 = 2010

Hence, the expression is equal to

Simplifying,

= x² – 5 = (2007)² – 5 = 4, 028, 044

When the polynomial x²⁰⁰⁷ + 1 is divided by x² – 3x +2, the remainder can be expressed in the form ax+b. Find the value of a.

My Solution:
Let k be the quotient, then:
x²⁰⁰⁷ + 1 = k(x² – 3x +2) + (ax+b)
Since x² – 3x +2 = (x-2)(x-1),
x²⁰⁰⁷ + 1 = k(x-2)(x-1) + (ax+b)

To omit k and to solve for the value of a easily, set x=2 and x=1:

For x=2
(2)²⁰⁰⁷ + 1= k(2-2)(2-1) + (2a+b)
2²⁰⁰⁷ + 1= 2a+b  (*)

For x=1
(1)²⁰⁰⁷ + 1 = k(1-2)(1-1) + (a+b)
2 = a+b  (**)

To solve for the value of a, perform elimination using (*) and (**).
..Which in turns give: a=2²⁰⁰⁷-1

Compute the number of squares between 4⁹ and 9⁴

My short solution: The technique is to express these two numbers into perfect squares so that it’s easy to compute the number of squares between these two numbers.

4⁹ = (2²)⁹ = (2⁹)² = 512² ;

9⁴ = (3²)⁴ = (3⁴)² = 81²

Therefore, the number of squares between 4⁹ and 9⁴ is (512-81)-1 = 430

Five digit numbers containing all digits from 1-5 are arranged from highest to lowest. What is the 25th number?

(Examples of these numbers are 54 321; 23 451; 41 532)

My Solution:

First, it’s impractical to list these (5)(4)(3)(2)(1)=120 numbers from highest to lowest and then determine the 25th number in the list. So, we can instead do some technique and investigate the distribution of these numbers.

Since it’s from highest to lowest, the first step is to determine the number of these five digit numbers which have a ten thousands digit of 5.

For thousands digit: 4 numbers to choose.

For hundreds digit: 3 numbers to choose.

For tens digit: 2 numbers to choose.

For unit digit: 1 number to choose.

Thus, the number of these 5 digit numbers which have a ten thousands digit of 5 is (4)(3)(2)(1)=24. Then these 24 numbers are the first 24 numbers in the list. Hence, the 25th number is the highest number which has a ten thousands digit of 4. And this 25th number is 45 321.

Prove that:    , for any

Solution: Since     , then in order to produce a short

proof, the expression   should be express in the form  

Doing so,

Hence,

At what angle of projection is the range maximum? Prove it mathematically.

To solve for the required angle of projection ( ), we need first to find a general expression for the range.

To find range, set y = 0 first and use:

,

Implies that,

(non zero value)

Solving for the general expression for range,

Regardless of the value of , the maximum value of range would be attain if is maximum. Since 1 is the maximum value of sine function, then, should be equal to 1. Since sin 90º = 1, therefore (angle of projection) should be equal to 45º

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