can anyone prove that if f(x)=0 then f(x) X g(x) = 0 for any function g? or is it jsut common sense.?

Answer 1

Lemma: Suppose a is any number. Then 0×a = 0.

Proof:

0 = 0 + 0
0×a = (0 + 0)×a
0×a = 0×a + 0×a
0×a - 0×a = 0×a + 0×a - 0×a
0 = 0×a + 0
0 = 0×a

QED =)

For any value of x, g(x) is a number.

Answer 2

they intersect the place the applications are equivalent. So set the two equivalent to a minimum of one yet another x^2 + 3x - 9 = 0.25x^2 subtract the 0.25x^2 over to get = 0 0.75x^2 + 3x - 9 = 0 .seventy 5 = 3/4 so I multiply with the help of four to sparkling denom 3x^2 + 12x - 36 = 0 reduces with the help of three x^2 + 4x - 12 = 0 (x + 6)(x - 2) = 0 x = -6 or x = 2 plug -6 and a couple of back in to the two function to discover the y factors of the coordinates. (-6, 9) and (2, a million)

Answer 3

Hi there,

The proof is that any number times 0 = 0.

If f(x) = 0, then it doesn't matter what g(x) is because 0 x any number = 0.

For example, if g(x) = 4:

f(x) x g(x) = 0

0 x 4 = 0 : So this proof must be true.

Answer 4

If you have two functions f and g defined on a set A
f:A--->B, g:A---->B
you can define another function:
h:A----->B

h(x) = f(x)*g(x)

In other words h = f*g in the set of functions.

Now if f(x) = 0 for all x in A, you have:

h(x) = 0*g(x) = 0, because any number multiplied by 0 is 0.

Therefore: h=f=0, i.e. h(x)=f(x)=0, for any x in A

Similarly, you can define:
k(x) = f(x)+g(x), k:A--->B
k=f+g

and if f(x)=0
k(x)=g(x) or k=g

Answer 5

anything times zero is zero. I don't know if this is only common sense, a mathematical definition or if there is a mathematical proof. A related question has to do with euclidean geometry vs non-euclidean geometry. In the first, parallel lines never intersect, in the other, they do. In math, things can get pretty strange when you explore more advanced concepts.

I hope this helps

Answer 6

But this is not always the case.

if f(x) = 0 then you can solve for x to equal something. Given that x-value, g(x) might be indeterminate or undefined. Therefore, their product would also be indeterminate or undefined. (with the exception of limit evaluations). Your assumption that it is an intuitive truth would be false.