## Nanomath – Proof

#### Proof

Some basics:

If n is an integer, consecutive integers could be either side i.e. n-1, n, n+1, n+2 etc.

Regardless of whether n is even or odd, 2n will be even, and 2n-1, and 2n+1 will be odd

Multiples of 2, 3 and 5 are written 2n, 3n, 5n respectively

##### “Prove algebraically that the sum of two even numbers is even”

Let m and n be two numbers, then 2m and 2n will be even numbers, so $2m + 2n = 2(m + n)$

As this is a multiple of 2, it is an even number

##### “Prove algebraically that the sum of two odd numbers is even”

Let m and n be two numbers, then 2m+1 and 2n+1 will be odd numbers, so $(2m + 1) + (2n + 1) = 2(m + n + 1)$

As this is a multiple of 2, it is an even number

##### “Prove algebraically that the sum of an odd and an even number is odd”

Let m and n be two numbers, then 2m+1 will be odd and 2n will be  even, so $(2m + 1) + (2n) = 2(m + n) + 1$ $2(m + n)$ is a multiple of 2, so it is an even number

Therefore, $2(m + n) + 1$ is an odd number

##### “Prove algebraically that the sum of three consecutive integers is divisible by three”

Let n be an integer, then n+1, and n+2 will be consecutive integers. $n + (n + 1) + (n + 2) = 3n + 3$ $= 3(n + 1)$

Since 3 is a factor of this result, so the sum of the 3 consecutive integers will be divisible by 3.

##### “Prove that the product of any three consecutive integers is divisible by 6”

In any 3 consecutive integers, one number is always a multiple of 2, and one will be a multiple of 3

Let the multiple of 2 be written 2n and the multiple of 3 be written 3m

Their product is (2n)(3m) = 6mn

Therefore, the product having a factor of 6, is divisible by 6.

##### “Prove algebraically that the difference between the squares of any two odd integers is divisible by 8”

Let two integer numbers be m and n, then 2m+1 and 2n+1 will both be odd numbers

Using the difference of two squares i.e. $a^2 - b^2 = (a+b)9a-b)$ $(2m + 1)^2 - (2n+1)^2 =[(2m+1)+(2n+1)][(2m+1)-(2n+1)]$ $=[2m+2n+2][2m-2n]$ $=2(m+n+1)2(m-n)$ $=4(m+n+1)(m-n)$

So 4 is a factor

If m and n are both even then (m-n) is even, and has a factor of 2

If m or n is even and the other odd then (m+n+1) is even and has a factor of 2

Therefore $(2m + 1)^2 - (2n+1)^2$ is divisible by 8

##### “Prove algebraically that the difference between the squares of any two consecutive integers is equal to the sum of the two integers”

Let n be an integer, then n+1 will be a consecutive integer. $(n + 1)^2 - n^2 = n^2 + 2n + 1 - n^2$ $= 2n + 1$ $= n + (n+1)$

This represents the sum of the two consecutive integers.

##### “Prove algebraically that the sum of the squares of any two consecutive numbers always leaves a remainder of 1 when divided by 4”

Let n be an integer, then n+1 will be a consecutive integer. $n^2 + (n + 1)^2 = n^2 + (n^2 + 2n + 1)$ $= 2n^2 + 2n + 1$ $2n(n+1)$ has a factor 2, and as n and n+1 are consecutive, one must be even, and therefore be a multiple of 2

So $2n(n+1)$  must be divisible by 4

Therefore $2n(n+1) + 1$ must have remainder 1 when divided by 4

##### “Prove algebraically that the sum of two consecutive multiples of 5 is always an odd number”

Let n be an integer, then 5n is a multiple of 5, and 5(n+1) will consecutively be the next multiple of 5 $(5n) + 5(n + 1) = 5n + 5n + 5$ $= 10n + 5$ $= 5(2n + 1)$

Whether n is even or odd, $2n + 1$ will be odd

Since 5 and $2n + 1$ are odd numbers, their product is an odd number

##### “Prove algebraically that the product of two consecutive multiples of 5 is always an even number”

Let n be an integer, then 5n is a multiple of 5, and 5(n+1) will consecutively be the next multiple of 5 $5(n + 1)(5n) = 25n(n + 1)$

Either n will be odd and n+1 will be even, or vice versa

Since 25 is odd, odd x odd x even = even, i.e.  their product is an even number 