I thought Id start a new exercise thread since nobody seems to be reading the first thread any more.
Let a[sub]1[/sub], a[sub]2[/sub], a[sub]k[/sub] be k consecutive terms of an arithmetic progression with common difference d (where the a[sub]i[/sub] are integers). Show that if gcd(k,d) = 1, then k divides a[sub]r[/sub] for some r = 1, , k.
Let a, b, c be three consecutive integers. Prove that at least one of 10a−1, 10b−1 and 10c−1, is not prime.
Last edited by JaneFairfax (2007-12-30 06:53:00)
uuhhh... nice,i guess LOL MAN THATS CONFUSING
Man, what a pain...
Yeah, I made both problems up myself. On reflection, I now realize that #1 is a bit on the tough side. Sorry.
Last edited by JaneFairfax (2007-11-06 00:25:32)
edit: forgot to hide it
Last edited by Kurre (2007-11-20 04:13:50)
Let f and g be real-valued functions which are twice differentiable on an interval I. Show that if f″(x) ≠ g″(x) for all x ∈ I, the graphs y = f(x) and y = g(x) cannot intersect each other more than twice in the interval I.
igloo myrtilles fourmis
a, b, c, d are real numbers such that 0 ≤ a ≤ b ≤ c ≤ d and a + d ≤ b + c. Show that ad ≤ bc.
ad ≤ bc
Nice work. (The last part of the proof needs a little bit of touching up, thata all.)
Let a, b, x, y be rational numbers. Consider the following statement.
(i) Give a counterexample to show that the above statement, as it stands, is not true.
(ii) What condition must be imposed on the above statement to make it true?
I had another method for 3 half-formed in my mind yesterday.
After a good night's sleep I've fortified it, so I thought I'd post even though I'm late.
Let p and q be the mean and half-range of a and d, and define r and s similarly for b and c. That is:
p = (a+d)/2
q = (d-a)/2
r = (b+c)/2
s = (c-b)/2
Then it is a simple matter to show that:
p-q = a
p+q = d
r-s = b
r+s = c
Therefore ad = (p-q)(p+q) = p²-q² and bc = (r-s)(r+s) = r²-s².
Further, using the fact that 0≤a≤b≤c≤d shows that c-b≤d-b≤d-a and so s≤q.
Also, since a+d≤b+c, then p≤r.
Hence p²-q² ≤ r²-q² ≤ r²-s² and thusly ad ≤ bc, WWWWW.
Why did the vector cross the road?
It wanted to be normal.
Nice solution, Mathsy.
Actually #3 wasnt exactly mine; it was something I copied from the Web, which I managed to solve myself. Anyway, the following exercise question is mine.
A cubic polynomialhas roots . Another cubic polynomial has roots . If the leading term in is , find (in terms of p, q and r) the coefficient of the term in in .
There is probably noone looking here, but anyway:P:
lets call the requested coefficient t.
according to vietes identity:
Using the inequality between arithmetic and geometric metdium yields:
letting ab=1/c, bc=1/a, ca=1/b, yields
1/c=2/a=4/b -> b=2a and c=a/2
abc=1 ->a*2a*a/2=1 -> a=1, which gives that b=2, c=1/2
nice problem btw
Last edited by Kurre (2008-05-16 01:59:06)
Suppose the roots of the cubic equationare all real and non-negative. Prove that
What condition is necessary and sufficient for equality to occur?
Its also possible to use the rearrangement inequality directly on the first inequality. However, I spotted an error on my solution to the second one:
Let R and S be rings with multiplicative identities 1[sub]R[/sub] and 1[sub]S[/sub] respectively and suppose f:R → S is a function which preserves multiplication (i.e. f(xy) = f(x)f(y) for all x, y ∈ R). Show that if f does not map all the elements of R into the zero element of S and S is an integral domain, then f(1[sub]R[/sub]) = 1[sub]S[/sub].