Very general/simple question.

Suppose a question asked something like this as an example:

1) given the set

\displaystyle \{ 0,1,2,3,4,5 \}

show that for any x in the set there exists a y in the set such that x*y=0.

Now clearly for 1,2,3,4,5 you can say let x=1,2,3,4,5 and y=0 but my question is this, is it allowed to say x=y=0 then x*y=0 and 0 is in the set. I guess I'm wondering if we can let x=y or would we need another 0 to justify it i.e. let x=0 (the first zero say) and y=0 (the other zero).

I hope I explained this well enough.

Thanks

Reply 1

davros

Original post by poorform

Suppose a question asked something like this as an example:

1) given the set

\displaystyle \{ 0,1,2,3,4,5 \}

What you've written makes no sense whatsoever!

What makes you think you can assume x = y?

Reply 2

rayquaza17

Original post by poorform

Suppose a question asked something like this as an example:

1) given the set

\displaystyle \{ 0,1,2,3,4,5 \}

If you can explain it for x=1,2,3,4,5 where y=0, then you can easily explain it for x=0 because it's the same argument with x and y interchanged.

(edited 9 years ago)

Reply 3

poorform

Okay let me try and explain in better example.

Suppose you were asked to prove/disprove the following statement.

For any

\displaystyle x \in \{-2,-1,0,1,2\}=M~\exists ~y \in M

such that

\displaystyle x+y=0

now if we choose

\displaystyle x= \pm 1 , \pm 2

then we can choose

\displaystyle y

to be the same number with a different sign e.g.

\displaystyle -2+2=0

now we have shown the statement is true for

\displaystyle x= \pm 1, \pm 2

now here is my issue to show the statement is true overall then we must show it is true for

\displaystyle x=0

now clearly if we add anything to

\displaystyle 0

other than itself we will get a non-zero answer so by showing this would the statement be false. Or is there any validity in saying choose

\displaystyle x=0

then

\displaystyle \exists ~y \in M s.t x+y=0

with this

\displaystyle y

being equal to

\displaystyle 0

as well or is this just utter rubbish along with the statement.

Reply 4

lizard54142

Original post by poorform

Suppose a question asked something like this as an example:

1) given the set

\displaystyle \{ 0,1,2,3,4,5 \}

If x = 0, then for any value of y x*y=0.
If x

\neq

0, then one of the remaining values must be zero, hence you can choose this for y and x*y=0

Reply 5

theOldBean

Original post by poorform

Okay let me try and explain in better example.

Suppose you were asked to prove/disprove the following statement.

For any

\displaystyle x \in \{-2,-1,0,1,2\}=M~\exists ~y \in M

such that

\displaystyle x+y=0

now if we choose

\displaystyle x= \pm 1 , \pm 2

then we can choose

\displaystyle y

to be the same number with a different sign e.g.

\displaystyle -2+2=0

now we have shown the statement is true for

\displaystyle x= \pm 1, \pm 2

now here is my issue to show the statement is true overall then we must show it is true for

\displaystyle x=0

now clearly if we add anything to

\displaystyle 0

other than itself we will get a non-zero answer so by showing this would the statement be false. Or is there any validity in saying choose

\displaystyle x=0

then

\displaystyle \exists ~y \in M s.t x+y=0

with this

\displaystyle y

being equal to

\displaystyle 0

as well or is this just utter rubbish along with the statement.

The following assertion is true:

Claim: For any

x \in \{-2,-1,0,1,2\}=M

there exists

y \in M

such that

x+y=0

Proof: We are told that

x \in M

, which, by definition of

M

, means that there are exactly five possibilities for the value of

x

. We check that the claim holds for each of these possibilities.
If

x=-2

then take

y=2

.
If

x=-1

then take

y=1

.
If

x=0

then take

y=0

.
If

x=1

then take

y=-1

.
If

x=2

then take

y=-2

.
In each case it is easy to see that

y \in M

and

x+y=0

, as desired.

I'm not sure if this is helpful to you - I don't really understand what your question is. Anyhow, there is nothing wrong with the proof I've written above, although in practise it's a bit laborious and I'd be tempted just to do the first three cases and say that the rest follows 'by symmetry'.

As DFranklin says (see the post below this), it is perfectly ok for x and y to take the same value.

(edited 9 years ago)

Reply 6

DFranklin

Original post by rayquaza17

x and y cannot be the same number.I might be missing something, but I really don't see why they can't be the same.

It's not very different from:

If G is a group, then for any x in G we can find y in G such that xy = e (identity).

In this case x and y absolutely can be the same element (most obviously, if x = e then we need to choose y = e as well).

Reply 7

Smaug123

Original post by poorform

Suppose a question asked something like this as an example:

1) given the set

\displaystyle \{ 0,1,2,3,4,5 \}

A set cannot contain duplicate elements. Let

X = \{0\}

. It's true that for every

x \in X

there is

y \in X

such that

xy=0

, and it is "the same zero" each time. It isn't possible to have "another zero to justify it".

If the question contained the words "with x and y distinct", then it would remain true; you'd just pick

y=1

for the x=0 case. You could no longer pick 0,0 because 0 is not distinct from 0; there's no way for the set to contain two non-distinct zeros. (That's the Axiom of Extensionality.)