Math Is Fun Forum

Haikus are sublime.
I can even make them rhyme
Though not all the time.

I think what the statement is saying is this. Choose any two positive integers a and b. Then (supposedly) a would be equal to b if max(a,b) = some positive integer n – so “prove” this by induction on n.

I don’t think there’s anything wrong with making such a statement. I mean, the statement is well and truly false; however, there is no harm at all in stating a statement of this kind – provided it can be shown that the supposed proof of it is invalid. And I think you’ve already exposed the fallacy – i.e. that while a and b may be positive integers, a−1 and/or b−1 may not be positive integers.

Viola is also a flower.

And it’s also a girl’s name. Viola is the heroine in Shakespeare’s play Twelfth Night and in Verdi’s opera La Traviata. (Actually, in the latter, the heroine is Violetta.)

Okay, here’s I(1).

From which we can see that

If we carry on, we find

From the emerging pattern, we can guess that

It remains to prove our guess by induction. That’s the easy part – the hard part is finding the pattern in the first place. Well, not so much hard as laborious.

The way to prove it – as Ricky has pointed out – is by a vacuous proof.

A statement of the form “if P then Q” is said to be vacuously true iff P is false. (It does not matter whether Q is true or not. As long as P is false, the whole of “if P then Q” is true.)

Now, the definition of “A is a subset of B” is: “For all x, if x is in A, then x is in B”. If you let A be the empty set and B be any set, you can clearly see that the statement “x is in A” is false for all x. Therefore the statement “if x is in A, then x is in B” is automatically true if A is the empty set. Hence A (the empty set) is a subset of B (any set).

(Note: Whether x is actually in B or not is immaterial. We are not interested in the truth value of “x is in B”; we are only interested in the truth value of “if x is in A, then x is in B”. And this is automatically (vacuously) true if “x is in A” is false.)

Or just

#2

(1) C is the circle of radius 1 with centre at (1,1). It touches (is tangent to) the y-axis at (0,1). So if M is this point and MP is perpendicular to MQ, PQ must be the diameter of the circle. Thus, the line ℓ passes through the centre of the circle – so k = 1.

EDIT:
(2) Still working on the second part. The relationship I’ve derived between b and k is

Wait! I just thought of another method, using complex numbers.

Consider the roots of the complex equation

The roots are

The sum of the roots must be 0. (In general, the sum of the roots of a polynomial equation of degree n is given by −(coefficient of z[sup]n−1[/sup]) divided by coefficient of z[sup]n[/sup].)

If you sum the roots, the imaginary parts of the complex conjugates cancel in pairs, leaving twice the real part, i.e.

Note that

And that’s the problem solved!

MathsIsFun started a thread about it:
http://www.mathsisfun.com/forum/viewtopic.php?id=6474

Do (b) first. You need (b) to prove (a)

For (a)

The result follows from (b).

Note: In any ring, −ab = (−a)b = a(−b). This can be proved from the distributive ring axiom and the fact that c·0 = 0·c = c for any element c in the ring. See this.