The web site of fmhelper

Unit 1
Unit 2
Unit 3
Unit 4
Unit 5
Unit 6
Email Us

Proofs

Information

Proofs are used in mathematics in order to shorten the time it takes to complete a future operation as by proving a method works to other mathematicians allows it to be widely used and can save everyone alot of time if it is a shorter method

Below is a simple prrof for finding if everything multiplied by 2 is an even number

First we can define a natuaral numbers as n

Therefore doubling n gives us 2n which is divisible by 2 therefore for n this is correct

Next we do the same operation for n+1 to see if the next number after n is also divisible by 2
2(n+1) = 2n + 2
This is also divisible by 2 and therefore the statement is also true for n+1

Finnaly we do this operation for 1
2×1 = 2
This is divisible by 2 as well and therefore we can state:

For all Natural Numbers multiplying by 2 gives an even number

Steps for a proof

First we see if true for a value of 1
Then we see if true for a value of a constant (e.g.n or k)
Next we check for the constant plus 1 (e.g.[n+1] or [k+1])
Finally we write a paragraph about what we have proved and how we have proved it

Example 1

Prove:

r=1n1r(r+1)=nn+1

Assume n=1:

r=1n1r(r+1)=11(1+1)=12

nn+1=11+1=12

Therefore this is true for n = 1 as both equations have the same answer
Now assume true for n = k

r=1k1r(r+1)=kk+1

Now consider for k+1:

r=1k+11r(r+1)=11×2+12×3+13×4+...+1k(k+1)+1(k+1)(k+2)

As we assumed that for n=k:

r=1k1r(r+1)=kk+1

We can rewrite the equation as:

r=1k+11r(r+1)=kk+1+1(k+1)(k+2)=k(k+2)(k+1)(k+2)+1(k+1)(k+2)

=k(k+2)+1(k+1)(k+2)=k2+2k+1(k+1)(k+2)=(k+1)2(k+1)(k+2)=k+1k+2

Now for the other equation:

k+1(k+1)+1=k+1k+2

We once again have the same answer from both equations for k+1 and it is therefore true for k+1 as well

(Must Right This)

But this is given the result with k+1 replacing k
Therefore, if the result is true for n=k it is also true for n = k+1
Since it is true for n=1
By induction it is therefore true for all positive integers n

Example 2

Prove that for all positive integers n,

1+2+3+4+...+n=(n2)(n+1)

First we test for n=1

1=(12)(1+1)=1

This is true and therefore the statement is true for n=1

Now we assume this is true for n=k

1+2+3+4+...+k=(k2)(k+1)

Letting n=k+1

1+2+3+4+...+k+(k+1)=(k2)(k+1)+(k+1)=k(k+1)2+2(k+1)2

=k2+k2+2k+22=k2+k+2k+22=k2+3k+22=(k+1)(k+2)2

Replacing k with k+1 in:

1+2+3+4+...+k=(k2)(k+1)

1+2+3+4+...+k+(k+1)=(k+12)([k+1]+1)=(k+12)(k+2)=(k+1)(k+2)2

This is the same value we got in the other equation therefore the statement is true for k+1

But this is given the result with k+1 replacing k
Therefore, if the result is true for n=k it is also true for n = k+1
Since it is true for n=1
By induction it is therefore true for all positive integers n

Example 3

Prove that for all positive integers n:
12+22+32+...+n2=(n6)(n+1)(2n+1)

Letting n=1

12=(16)(1+1)(2[1]+1)=(16)(2)(3)=1

This is true and therefore the statement is true for n=1

Now we assume this is true for n=k

12+22+32+...+k2=(k6)(k+1)(2k+1)

Letting n=k+1

12+22+32+...+k2+(k+1)2=(k6)(k+1)(2k+1)+(k+1)2

Taking common parts outside the brackets

=k+16[k(2k+1)+6(k+1)]=k+16[(2k2+k)+(6k+6)]

=k+16(2k2+7k+6)=k+16(k+2)(2k+3)

Replacing k with k+1 in:

12+22+32+...+k2=(k6)(k+1)(2k+1)

12+22+32+...+k2+(k+1)2=(k+16)([k+1]+1)(2[k+1]+1)

=(k+16)(k+2)(2k+3)

This is the same value we got in the other equation therefore the statement is true for k+1

But this is given the result with k+1 replacing k
Therefore, if the result is true for n=k it is also true for n = k+1
Since it is true for n=1
By induction it is therefore true for all positive integers n

Example 4

Now we will do a different kind of proof where we are proving we can find a number in a sequence using an equation:

Given:

u1=2,un+1=4un+3

Prove:

un=3×4n11

First we test for n=1

u1+1=4u1+3

First we test for n=1

u2=4(2)+3=11

Now testing:

un=3×4n11

u2=3×4211=3×411=121=11

This is true and therefore the statement is true for n=1

Now we assume this is true for n=k

uk+1=4uk+3

and

uk=3×4k11

Letting n=k+1

uk+1=3×4(k+1)11=3×4k1

Replacing k with k+1 in:

uk+1=4uk+3

Using the value of

uk

we have already found:

uk+1=4(3×4k11)+3=4(3×4k11)+3=(4×3×4)k14+3=(4×3×4)k11

Now we need to remember that

41=14

and therefore

4k1=14(4k)

Employing this knowledge to set the power to k in our equation from k-1 gives:

uk+1=14(4×3×4k)1=3×4k1

Now we have the same value that we got previously and therefore the statement is true for k+1

But this is given the result with k+1 replacing k
Therefore, if the result is true for n=k it is also true for n = k+1
Since it is true for n=1
By induction it is therefore true for all positive integers n

Example 5

Prove

2n+2+32n+1

is divisible by 7 for all values n ≥ 1

First we let n=1

21+2+32(1)+1=23+33=8+27=35

Now we need to check that 35 is divisible by 7

357=5

This is divisible by 7 therefore this statement is true for n=1

Now we assume this is true for n=k

2k+2+32k+1=7m

Where m is any integer value (Therefore 7m is always divisible by 7)

We can rearrange this to:

32k+1=7m2k+2

Letting n=k+1

2(k+1)+2+32(k+1)+1=2k+3+32k+3=2×2k+2+32×32k+1

Replacing

32n+1

with

7m2k+2

from the equation derived above gives us:

(This can be done with the

2k+2

as well if you rearrange the equation differently)

2×2k+2+32×32k+1=2×2k+2+9(7m2k+2)

=2×2k+2+63m9×2k+2=63m7×2k+2=7(9m2k+2)

This is a multiple of 7 and is therefore divisible by 7 meaning that the expression is true for n=k+1

But this is given the result with k+1 replacing k
Therefore, if the result is true for n=k it is also true for n = k+1
Since it is true for n=1
By induction it is therefore true for all positive integers n

Example 6

This final example shows a proof for a matrix

For

A=(12 02)

,Prove that

An=(12n+12 02n)

Letting n=1

A1=A=(121+12 021)=(1222 02)=(142 02)=(12 02)

This is the same value we have for A and therefore it is true for n=1

Now we assume this is true for n=k

Ak=(12k+12 02k)

Letting n=k+1

Ak+1=(12(k+1)+12 02k+1)=(12k+22 02k+1)

We know that

A×Ak=Ak+1

,Therefore using this multiplication we can see if we get the same equation for

Ak+1

A×Ak=(12 02)(12k+12 02k)=(12k+22 02k+1)

This is the same expression we found for n=k+1 meaning that the expression is true for n=k+1

But this is given the result with k+1 replacing k
Therefore, if the result is true for n=k it is also true for n = k+1
Since it is true for n=1
By induction it is therefore true for all positive integers n