onto functions?

Question

im wondering if the following functions are onto or not...surjective

f(m,n)=2m - n
f(m,n)=m^2 - n^2
f(m,n)=m+n+1
f(m,n)= |m| + |n|
f(m,n)= m^2 - 4

f: Z X Z ----> Z

Answer 1

"onto" only makes sense with respect to a range. If you specify the range (like integrers, natural numbers, reals, etc) and the domain of m and n, then "onto" makes sense.

Ok, since the domain and range are Z,
f(m,n)=2m - n
onto since f(n,n) maps to n for any n in Z
f(m,n)=m^2 - n^2
not onto, you cannot get the number 2
f(m,n)=m+n+1
onto, f(m,-1) maps to m
f(m,n)= |m| + |n|
not onto, cannot get negatives
f(m,n)= m^2 - 4
not onto, cannot get numbers less than -4 or numbers or most positive numbers for that matter

Answer 2

First, get a respectable instinct about the function, and the way it quite works. enable's write out some words, in the type of a series: h(0), h(a million), h(2), h(3), h(4), h(5), h(6), ... 0, 2, 3, 5, 6, 8, 9, ... it kind of feels to omit one in 3 numbers. to reveal that it truly is no longer onto, we in reality favor to reveal it won't be able to benefit one decision (now to not indicate infinitely many). enable's %. a million. How do we educate that no h(x) exists such that h(x) = a million? i imagine the most intuitive way is to word that h(x) is a non-lowering function. both the ceiling and floor applications are non-lowering, in that: x >= y ===> ?x? >= ?y? x >= y ===> ?x? >= ?y? this should be really sparkling. Then, to reveal that h(x) is non-lowering, we think x >= y. Then: ?x? >= ?y? ?x? >= ?y? ===> 2?x? >= 2?y? at the same time, we get: ?x? + 2?x? >= ?y? + 2?y? i.e. h(x) >= h(y). Now, how does this help? properly, for the reason that h(0) = 0, it follows from this resources that for each x such that x <= 0: h(x) <= h(0) = 0 < a million so no h(x) = a million for x <= 0. in the different case x > 0, meaning that x >= a million. as a effect, by a similar resources: h(x) >= h(a million) = 2 > a million So no x by any ability will fulfill h(x) = a million. this signifies that h isn't onto. wish that permits!

Answer 3

Onto what? N²→N? Z²→Z? R²→R? The answer depends on this.

For example, the last case
f(m,n)= m^2 - 4

is R² onto R because f(sqrt(x+4), 0) = x for all x E R. But it is not N² onto N because not all natural numbers are obtained by removing 4 from a square.

Answer 4

The other guys took it as far as they could without your extra information (f: Z X Z ----> Z). Just in case they don't come back, I'll throw in an answer. Hope it's right.

f(m, n) = 2m - n

Pretty sure that's onto. f(0, -x) = x.

f(m, n) = m² - n²

I don't think this is onto. For instance, I don't believe there are two perfect integer squares whose difference is exactly 2.

f(m, n) = m + n + 1

Again I'm sure pretty this is onto. f(x - 1, 0) = x

f(m, n) = |m| + |n|

This is definitely NOT onto, as it's impossible to get a negative integer output.

f(m, n) = m² - 4

This one would only allow integers that are 4 less than a perfect square, so you couldn't get 6, for instance. So I don't think it's onto.

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Also, vacaloco thumbs-downed us all so his answer would appear to be better.

Answer 5

f(m,n)=2m - n
yes.
f(m,n)=m^2 - n^2
no, for example 2 does not have a preimage
f(m,n)=m+n+1
yes
f(m,n)= |m| + |n|
no, because f(m,n) >=0, so the negative numbers do not have a preimage.
f(m,n)= m^2 - 4
no, because m^2 -4>=-4, so any number smaller than -4 will not have a preimage .