Bite-size chunks of incredibly important mathematics
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
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
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
is a multiple of 2, so it is an even number
Therefore,
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.
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.
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
is divisible by 8
![(2m + 1)^2 - (2n+1)^2 =[(2m+1)+(2n+1)][(2m+1)-(2n+1)]](https://s0.wp.com/latex.php?latex=%282m+%2B+1%29%5E2+-+%282n%2B1%29%5E2+%3D%5B%282m%2B1%29%2B%282n%2B1%29%5D%5B%282m%2B1%29-%282n%2B1%29%5D&bg=ffffff&fg=000000&s=0&c=20201002)
![=[2m+2n+2][2m-2n]](https://s0.wp.com/latex.php?latex=%3D%5B2m%2B2n%2B2%5D%5B2m-2n%5D&bg=ffffff&fg=000000&s=0&c=20201002)
“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.
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.
has a factor 2, and as n and n+1 are consecutive, one must be even, and therefore be a multiple of 2
So
Therefore
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
Whether n is even or odd,
will be odd
Since 5 and
“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
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
