Download Objective: using algebra to prove number facts 1. Prove that the

Objective: using algebra to prove number facts
1. Prove that the square of any odd number is always one more than a multiple of 8.
2. Show that the sum of any three consecutive multiples of 3 is also a multiple of 3.
3. Write down the nth term of the sequence 4, 7, 10, 13, 16, … Prove that the product of any two
terms of this sequence is also a term of the sequence.
4. In this question a and b are numbers where a = b + 2. The sum of a and b is equal to the product
a and b. Show that a and b are not integers.
5. Show that (2a – 1)2 – (2b – 1)2 = 4(a – b)(a + b – 1). Prove that the difference between the
squares of any two odd numbers is a multiple of 8.
6. Prove that (3n + 1)2 – (3n – 1)2 is a multiple of 4, for all positive integer values of n.
Prove that the square of any odd number is always one more than a multiple of 8.
1.
The square of an odd number can be written as  2n  1  4n2  4n  1
2.
We have to show that this is one more than a multiple of 8. This does not look obvious, although we
have the + 1, 4n2  4n  1 , it does not seem as though 4n2  4n is a multiple of 8.
3.
But 4n2  4n  4 n2  n . So if n was even then n 2  n would also be even, and so we would an
2




even number multiplied by 4 which is a multiple of 8.
4.


If n was odd then we would have to consider that for n 2  n , an odd number squared is odd plus
an odd number is even. So again we have an even number multiplied by 4 which is a multiple of 8.
1.
Show that the sum of any three consecutive multiples of 3 is also a multiple of 3.
3n   3n  3   3n  6   9n  9
2.
But we have to show that 9n  9 is a multiple of 3, i.e. of the form 3
3.
So factorise 9n  9  3  3n  3
4.
Since  3n  3 is any whole number. Then to multiply it by 3 gives a multiple of 3.
1.
2.
Write down the nth term of the sequence 4, 7, 10, 13, 16, …
Prove that the product of any two terms of this sequence is also a term of the sequence.
nth term of the sequence is: 3n  1
We are asked to show that the product of any two terms of this sequence.
Therefore it will not be right to start with  3n  1 3n  1 as these are identical
terms of the sequence. So introduce a new variable.  3n  1 3k  1
3.
OUR GOAL is to show that the product is a term of the sequence, i.e. of the form 3n  1 or more
generally 3
1
4.
So manipulate  3n  1 3k  1
5.
 3n  13k  1  9nk  3n  3k  1
6.
3
7.
3  3nk  n  k   1 , as required.
1.
In this question a and b are numbers where a = b + 2. The sum of a and b is equal to the
product a and b. Show that a and b are not integers.
We are told that the sum a and b is equal to the product a and b . That is
a  b  ab
Since we know that a  b  2 let us substitute this into the equation a  b  ab so that we have only
2.
 1 , we need to take out a factor of 3 and ensure that we have the plus one.
one variable.
3.
4.
b  2  b   b  2 b
b  2  b  b  2 b
2
This becomes 2b  2  b  2b
2  b2
integer too as a  b  2
and we know that square root of 2 is not an integer, so a is not an