The way we teach math and sciences has been behind technology development for years. With Wolfram Alpha, many traditional educators and even parents will protest, as if the students were not learning. However, I believe the education system is chasing after the Wolfram Alpha style. Just because a website can find complete background information on a math function, it does not mean there is nothing left to teach. Educators need to become more creative and education itself needs to become more in-depth.

Wolfram alpha will be bad for _traditional_ calculation based approaches to math teaching. Students will be tempted to use it for homework problems, then not have the experience to pass the exams. Thus, WA will force a different, and hopefully better, approach to teaching math. (of course these comments only apply to high school and undergrad courses, where calculation is dominant)

A nice intro into and discussion of W|A impact on the math education could be found on Walpha Wiki site:
http://walphawiki.wikidot.com/

hello, i have honor to writing for you my article that explaine my new basic(inequality),and all thanks for ur encourage of publisher mathematicien
this

is my article ( new inequality)

In the first I search for true hypothese to solved this
Probleme, we have for n=1,we have x+y=z
For x,y,z€R POSITIF, and
x≤z and,y≤z quel que soit x and y positive number reel
so we can get |x|≤|z|,and |y|≤|z |
so -|z|≤ x≤|z|……1
-|z|≤ y≤|z|…….2
I add the too i thus get
-2α≤x+y≤2α that α=|z| so α≥0…..3
In this cas x+y mayb will 2α-k that k≥0
Or 2α+k , k≤0 ,so
{x+y=2α-k if k≥0 or
{x+y=2α+k if k≤0
So -2α≤2α-k≤2α, so we can get
-3α≤α-k≤α but we have α=|z| ,and z≤ |z| ,z≥0
So z=α-k ,k≥0 , now we find x+y,and z, so
y=α+k
now I find the only hypothese to solve the fermat equation,
for n=1,I put x=α,y= α+k, z= α-k i remplaced with this value in the equation we find ; α-2k=0, so α=2k, and x,y,z after substitution we can find , x=2k,y=k,z=3k , this solution of x+y=z
and now I solve the equation for n=2, so it must be to solved this equation pythagorthien
x²+y²=z²
so I put x=α,y =α+k,and z= α-k after substitution in the equation of second degree we find, α=4k,so
x=4k,y=3k,z=5k, k€R
now I solve the equation with n=3,but its equation 3 degre it diffuclt
to solved,in this cas it must be used the perception
-whene we compare with precident solution for n=1,n=2, we find good idea, the defference of the solution for n=1,and n=2 is constant and equal 2k so,I can find the result and solution for n=3
So x=6k,y=5k,z=7k, but this solution verified by 0 yet,and whene we put k=1,so 6^3+5^3 not equal 7^3, the defference between tow is 2,so 2k, in this cas I propose opened sphere ,such with r=1/n, n defferent for 0, and 1/n € ]0,1[ this interval ouvert belong in v, the voisinage of sphere , and we can doing the subdivision of ]0,1[, in lot area
Such; it was find r’=j*1/n, but with observation I look the number
0.1 verified the equation x^3+y^3=z^3 and whene we substate in the general solution for n=3, we find 0.6^3+0.5^3=0.7 ^3
And its equal 0.34 in the tow tom of equation,in add to i chose
k= r’=j*1=0.1/n.
-for n=4 we find x=8k,y=7k,z=9k,so for n we can obtien the general
Relation, so for n ≥4, we find the solution like this
X=2nk,y=2nk-k,z=2nk+k,k belong in R number
So the solution of probleme fermat like this
And x^n+y^n=z^n it equivalent for this equation or solution
(0.2)^n+[(0.2) -(0.1/n)] ^n ≤ (0.2) +(0.1/n)] ^n
For n≥4, but 0.1 is it solution for n=3, so this is a last my solution
this is general basic
quel que soit n≥1
a, b ;reel number no nul and
a €]0,1[ b=a/2n
a^n+(a-b)^n ≤ (a+b)^n ……(1)
this is last general basic(1)
tel/00213697213515
all thanks,

Dear Sir
I would appreciate very much if you can
help me with the below question.
Thanks very much
Lawrence
n! = 3 to the power of 2007 find n

My e-mail is smsltd@starhub.net.sg

Thus, WA will force a different, and hopefully better, approach to teaching math.