NCETM_Mathematics_Departmental_Workshops_Index_Numbers_Resource_Sheet.HT2.FRA.3

The Equivalence of Roots and Fractional Powers

Using Pythagoras’ Theorem, we find that the diagonal of the triangle is… 1 1

2 Using Pythagoras’ 1 Theorem, we find that the 1 diagonal of the triangle is… 2

Construct a square on this diagonal. 2 1 1

Construct a square on this diagonal. 1 1

Divide it up into triangles. 1 1

Divide it up into triangles. 1 1

Each triangle has an area of 1 2 1 1

11 Each triangle 22 has an area of 11 1 22 2 11 2 1

So the square has an area of… 11 11 2 22 1 11 22

So the square has an area of… 2 2 1 1

Remember the length of the diagonal: 2 2 12 1

Remove the rest of the picture… 2 12 1

Remove the rest of the picture… 2 2

If it helps you, rotate the square a little bit… 2 2

2 If it helps you, rotate the 22 square a little bit… …that’s better!

2 This square reminds us that 22 2× 2=2

2x I wonder if we can do this 22x using powers of 2 instead of 2x × 2x = 2 roots…?

2x Let’s use the Index Laws to 22x investigate the algebra. 2x × 2x = 2

2x × 2x = 2 Let’s use the Index Laws to 2x × 2x = 21 investigate the x + x =1 algebra. 2x = 1 x= 1 2

2x × 2x = 2 Returning to our square… 2x × 2x = 21 x + x =1 2x = 1 x= 1 2

2 1 So we can see clearly that 22 22 1 21 2 = 22 22

We could have started with a square of any size. In general, then, we have: 1 N2 = N

Let’s look at this cube now. Imagine it has a volume of 2.

2

So its edge length must be the cube root of 2. 2

32 So its edge length must 32 be the cube root of 2. 23 2

32 In other words, 32 32× 32× 32 =2 23 2

32 Can we write the same 32 thing using powers of 2 instead of roots? 23 2 32× 32× 32 =2

2x Can we write the same 2x thing using powers of 2 instead of roots? 22x 2x × 2x × 2x = 2

2x Let’s use the Index 2x Laws as before. 22x 2x × 2x × 2x = 2

2x × 2x × 2x = 2 Let’s use the Index Laws as before. 2x × 2x × 2x = 21 x + x + x =1 3x = 1 x=1 3

2x × 2x × 2x = 2 Returning to the cube… 2x × 2x × 2x = 21 x + x + x =1 3x = 1 x=1 3

2x Returning to the 2x cube… 22x 2x × 2x × 2x = 2

1 1 Returning to the cube… 23 23 111 1 2 23 × 23 × 23 = 2 23

1 1 So we see now that 23 23 1 1 2 23 = 3 2 23

32 So we see now that 32 1 23 2 23 = 3 2

We could have started with a cube of any size. In general, then, we have: 1 N3 = 3 N This can be extended to fourth, fifth and higher roots too.