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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 1990 question 3 solution

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i) Given:

(1):

(2):

(3): .

From (1) and (2), by mutliplying by and , respectively, it follows

(4)

(5) .

Multiplying (3) by from the left and from the right, we have

and then subbing in (4) and (5) into RHS gives

as required.

ii) Assume that the orders of and are and , respectively.

Then,

(1)

Multiplying by from the left and by from the right gives

from which it follows that

.

But by using the exact same argument, only swapping c and d, we get that

from which it follows that

as required.

iii)

Given:

(1)

First part: Induction on s:

Statement:

(2)

The case s = 1 is merely the given (1).

Assume (2) is true for s = k, i.e.

Left multiplying by (1):

i.e. the statement is true for s = k + 1 as well, and thus is true by induction for all natural s. By instead using induction backwards, multiplying by ( is the inverse of ), we can show it to be true for all integers s.

Second part:

Known (from the last part): For any integer m,

(1)

Now use induction on n:

Statement:

Assume true for n = k:

Now let and (using (1)):

which proves the statements is true for n = k + 1 as well, and by induction blah blah..., as required.

Solution by ukgea.