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MATH 370: Homework 1
Due Thursday, 2/2 in class
Reading: Ross 1.1, 1.2
p
√
1. Show that 2 + 3 is irrational.
Let x be this real number. Then x satisfies the equation (x2 − 2)2 = 3. Hence, the
real number x is a root of the polynomial x4 − 4x2 + 1. Since neither ±1 are roots of
this polynomial, the theorem on rational zeroes of a polynomial with integer coefficients
implies that the real number x is irrational.
2. Find all rational solutions to 2x3 − x2 + 2x − 1 = 0.
This polynomial factors as (2x − 1)(x2 + 1), from which is follows easily that x = 1/2
is the only rational (in fact, only real) solution, but let’s pretend we didn’t know that.
By the rational zeroes theorem, the only possible rational solutions are ±1 and ±1/2.
Among the four possibilities, only 1/2 is a solution.
3. (a) Determine which of the axioms A1-A4, M1-M4, DL, O1-O5 (pp.13-14) fail for N. (b)
Which of these fail for Z ?
(A3), (A4) and (M4) fail for N (because 0 6∈ N, −1 6∈ N, 1/2 6∈ N, respectively) while only
(M4) fails for Z (because 1/2 6∈ Z).
4. Use induction to prove
|a1 + a2 + · · · + an | ≤ |a1 | + |a2 | + · · · + |an |.
Equality holds when n = 1, while for n > 1
|a1 + a2 + · · · + an | = |(a1 + a2 + · · · + an−1 ) + an |
≤ |a1 + a2 + · · · + an−1 | + |an |
≤ |a1 | + |a2 | + · · · + |an−1 | + |an |
= |a1 | + |a2 | + · · · + |an |.
(Remark: We showed more steps than necessary for pedagogical reasons.)
5. Assuming the Product Rule
dg
df
d
(f (x) · g(x)) = f (x) ·
+
· g(x)
dx
dx dx
to prove the Power Rule
d n
(x ) = nxn−1
dx
for all natural numbers n by induction.
For n = 1, the Power Rule follows from
d
x+h−x
(x) = lim
= 1 = x0
h→0
dx
h
while for n > 1, we have
d n
d
(x ) =
(x · xn−1 ) = x · (n − 1)xn−2 + xn−1 = nxn−1 .
dx
dx
6. Use induction to prove
12 + 22 + · · · + n2 =
n(n + 1)(2n + 1)
.
6
Both sides equal one when n = 1, while for n > 1,
12 + · · · + n2 =
n(n − 1)(2n − 1)
n
n(n + 1)(2n + 1)
+ n2 = (2n2 + 3n + 1) =
.
6
6
6
7. Use induction to prove that 7n − 6n − 1 is divisible by 36 for all natural numbers n.
This follows because 7n+1 − 6(n + 1) − 1 = 7(7n − 1) − 6n = 7(7n − 6n − 1) + 36n is
divisible by 36 as soon as 7n − 6n + 1 is, and because 71 − 6 − 1 = 0 is divisible by any
natural number, 36 in particular.
FOOD FOR THOUGHT: How is the Power Rule
d α
(x ) = αxα−1
dx
(x > 0)
for all real numbers α established?
It is possible to establish the Power Rule first for all rational α by elementary principles, then
extending the rule to all real numbers by using the density of rationals, but this approach is
quite tedious as many familiar fact about exponents need to be established from first principles.
The standard approach is to define xα as exp(α log x) where exp : R → R+ is the exponential
function and log : R+ → R is its inverse, a.k.a. the natural logarithm. The Power Rule is
more efficiently derived using the Chain Rule together with basic properties of the exponential
function and the natural logarithm. This approach is used even when the parameters x and α
are complex numbers, although the natural logarithm behaves more naturally as a multi-valued
function in this context.