Is this a valid maths proof?

A 'proof' needs to explain the working and your post just has statements, which may or may not be correct.

Unless I'm mistaken, the statements aren't even correct, as 'd' can equal 0.

see 'proof' below:

constants: a=0, b=0, c=0

d = 0 / ([x + 0] ^ 0)

=> d = 0 / x^0
=> d = 0/1
=> d = 0

=> d can be equal to 0 :p
 
No, if you let a = 0 then d will also be zero, regardless of the other values. You would have to state that 'a' is a non-zero number first.

It's also worth noting that even if it had been a 'true' statement, it still wouldn't have been a valid proof.

For it to be a mathematical proof, there would have to be much more detail as to how the end statement (in this case 'd cannot equal 0') had been deduced.
 
If a is non-zero, d cannot equal zero.

The limit of a / [x + b] ^ c as x->infinity is zero.

However, you're not really proving anything. All you have said is that you cannot have a quotient equal to zero if the numerator is non-zero, which is a well known axiom.
 
A proof would be something like:

(Using E as the symbol for set membership, and V for "for all", R is the set of real numbers)

Define: d(x) = a/(x + b)^c, where a,b,c are constants E R and where x E R.

Hypothesis: d /= 0

Proof:

Let a = 0, then d(x) = 0

Also

Take limits as x -> +- infinity

d(x) = lim x->+-inf a/(x+b)^c = 0.

Hence by contradiction d is not non-zero for some a,b,c,x E R.

However if constraints are imposed on a and x d(x) can be non-zero.

Let a E R/{0}, and x be finite:

d(x) = a/(x+b)^c

by observation d(x) can only = 0 if the numerator = 0 or the denominator = +- infinity, as a is non-zero and x is finite and b,c are constants those conditions cannot be satisfied. The condition x+b /= 0 must hold in order for the function to be defined.

And so d(x) /=0 V a,b,c,x E R under those constraints.

That was my attempt at a proof. Its probably not very acceptable at most levels.
(I've noticed some typos please don't execute me for those.)
 
Last edited:
However if constraints are imposed on a and x d can be non-zero.

Let a E R/{0}, and x be finite:

d(x) = a/(x+b)^c

by observation d can only = 0 if the numerator = 0 or the denominator = +- infinity, as a is non-zero and x is finite and b,c are constants those conditions cannot be satisfied. The condition x+b /= 0 must hold in order for the function to be defined.

And so d /=0 V a,b,c,x E R under those constraints.

I wouldn't say that's too much better than the op as a proof as you haven't justified why d can only = 0 if the numerator = 0 or the denominator = +- infinity.

A better and more elegant way would be to prove by contradiction:

Assume a /= 0 and d = 0 and x, b and c are real

then the equation can be rearranged as

d (x + b)^c = a

but left hand side = 0 * (x + b)^c

and right hand side = a /= 0

so by contradiction if a /= 0 then d /= 0.
 
Back
Top Bottom