STEP I 2008 Question 1 Help

What does it mean to say that a number x is irrational?
Prove by contradiction statements A and B below, where p and q are real numbers.
A: If pq is irrational, then at least one of p and q is irrational.
B: If p + q is irrational, then at least one of p and q is irrational.
Disprove by means of a counterexample statement C below, where p and q are real numbers.
C: If p and q are irrational, then p + q is irrational.
If the numbers

e, \pi, \pi ^2, e^2, e \pi

are irrational, prove that at most one of the numbers

\pi+e,[br]\pi - e, \pi^2 - e^2, \pi^2 + e^2

is rational.

I'm stuck as where to start with this question.

Reply 1

marty.frmn

just start putting a bunch of rational and irrational numbers for p and q and see what results u get in compliance with the statements given....

Reply 2

rbnphlp

11

Small123

What does it mean to say that a number x is irrational?
.

It cant be expressed as a fraction say m/n where m and n are integers..

Reply 3

Small123

OP

15

rbnphlp

It cant be expressed as a fraction say m/n where m and n are integers..

I get that but I can't think of any contradictions.

Reply 4

DFranklin

18

I'll do the first one for you.

A: If pq is irrational, then at least one of p and q is irrational.

Suppose this result is false, so that we can find two rational numbers p and q such that pq is irrational. But by definition, we can write p = a/b, q = c/d for some integers a,b,c,d.
Then

pq =\frac{ac}{bd}

, and so pq is actually rational. Contradiction.

Reply 5

Small123

OP

15

DFranklin

I'll do the first one for you.

A: If pq is irrational, then at least one of p and q is irrational.

Suppose this result is false, so that we can find two rational numbers p and q such that pq is irrational. But by definition, we can write p = a/b, q = c/d for some integers a,b,c,d.
Then

pq =\frac{ac}{bd}

, and so pq is actually rational. Contradiction.

Oohhhh, I get it! Thanks, I'll try the next one.
Suppose we can find two rational numbers p and q such that p+q is irrational. p=a/b, q=c/d where a,b,c,d are integers. Then

p+q=\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{ad+bc}{bd}

which is a contradiction. Is that right?

Yes.

OP

Any tips for C? I've had a few ideas which didn't quite work out as planned.

Reply 8

Small123

OP

15

Of Course! How did I not think o that :o:

. Thanks

Reply 9

TomP1806

For C you can use pi and (1-pi). Both irrational but they sum to make one, which is rational.

The last part is a bit more complicated, but i'll try my best to explain:

The question wants you to prove that at most one is ratioanl...so at least 3 are irrational.

Firstly:

1)

(\pi+e) + (\pi-e)

= 2pi...from A, at least one of these numbers must be irrational.

2)

(\pi^2+e^2) + (\pi^2-e^2)

=2pi^2...from A, at least one of these numbers must be irrational

Now assume

(\pi^2-e^2)

is rational. From 2) above, this must mean that

(\pi^2+e^2)

is irratioanal. Also,

(\pi^2-e^2)

=

(\pi+e) * (\pi-e)

. This means that either both

(\pi+e) * (\pi-e)

are rational, or they are both irrational, otherwise they could not form a ratioanl prodcut. From 1) they cannot both be rational as at least one must be irrational. Therefore they are both irratioanal.

Hence, if

(\pi^2-e^2)

is rational, there are at least 3 numbers which are irrational.

Assume then that

(\pi^2-e^2)

is irrational.

(\pi^2+e^2)

can be either irratioanal or rational. If it was irrational, then we'd have

(\pi^2-e^2)

assumed irrational,

(\pi^2+e^2)

as irrational and at least one of

(\pi+e) and (\pi-e)

as irrational...so at least 3 numbers are irrational.

Keeping with our assumption that

(\pi^2-e^2)

is rational, we will now assume that

(\pi^2+e^2)

is rational.

This must mean that

(\pi^2+e^2) + 2e\pi

and

(\pi^2+e^2) - 2e\pi

are both irrational

(\pi^2+e^2) + 2e\pi

=

(\pi+e)^2

(\pi^2+e^2) - 2e\pi

=

(\pi-e)^2

In both of these,

(\pi+e) + (\pi-e)

must either be both rational or irrational. From A, they cannot both be rational, so they must both be irrational.

Hence at least 3 are irrational.

Therefore whatever combination of rational/irratioanl of

(\pi^2+e^2) and (\pi^2-e^2)

you have, there will always be at least 3 irrational number => at most one ratioanl number.

Hope that kind of makes sense :smile:

STEP I 2008 Question 1 Help

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