Maths Marathon | Forum profile

Peanos axioms as I've been taught do not have a zero element. The integers arise from equivalence classes of pairs of natural numbers. Consider the equivalence relation (a, b) ~ (c, d) if and only if a + d = b + c. Under this relation, the element (a, a) acts as a zero element since (a, a) ~ (c, d) if and only if (a + d = c + a. Commutativity and cancellation give that d = c. The difficulty here is that there are many choices for (a, a),...

I've always learned the Peano axioms as follows: $\mathbb{N}$ refers to the natural numbers 1) For each $x\in\mathbb{N}$, $x=x$. 2) For all $x,y,z\in\mathbb{N}$, if $x=y$ and $y=z$ then $x=z$. 3) For all $x,y\in\mathbb{N}$, if $x=y$ then $y=x$. 4) If $x\in\mathbb{N}$ and $x=y$ for some $y$, then $y\in\mathbb{N}$. 5) 0 is an element of $\mathbb{N}$. 6) There exists a function $S:\mathbb{N}\to\mathbb{N}-\{0\}$ which is bijective 7) If $K$...

Technically, speaking, I think you could use division, since, for any a, b c, rational, the property holds, and the integers are a subset. This is circular, since cancellativity of multiplication is used to prove that the integers are a subset of the rational numbers, or that the rational numbers are well-defined.

Here's the questions, by the way: http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2009.pdf Looks like it was a (relatively) easy test this year. Even A6 isn't hard (though the number 6 may have prevented people from attempting it).

Claim: a + c = b + c ==> a = b Proof: a + 1 = b + 1 ==> S(a) = S(b). Since S(n) is injective, a = b. Suppose a + n = b + n . Then a + (n + 1) = b + (n + 1) ==> (a + n) + 1 = (b + n) + 1 ==> S(a + n) = S(b + n) ==> a + n = b + n by the injectivity of S(n), which is true by assertion. The first bolded statement is not what you want to assume. You want to assume that a + n = b + n ==> a = b. Then, a + (n...

I think I got 4 (A1, A2, B1, B2) right, though I handwaved a bit on B2... I may also have gotten some partial credit on B3 and/or B4. So basically I did better than I expected to do, but not as well as I could have.

Ah. I spent a little more time thinking about this tonight, and it seems not too hard to show not only that the set of continuity points of a function is G_\delta (which is the Baire Category argument I had in mind to rule out the rationals, since they're dense and meager), but also that every G_\delta is the set of continuity points of some real-valued function (essentially the same idea as in the counterexample (b) you originally asked...

I take it this year and I know we do Taylor Series with multiple variables.

"I want to make more money than my current degree allows me to, so let me in, or else!"

The rational root theorem basically says if a polynomial of the form a_nx^n + ... + a_0 has a rational root, then it will be of the form +p/q or -p/q where p is a factor of a_0 and q is a factor of a_n.