# don't understand part of answer to contrapositive question

• Sep 8th 2011, 08:46 PM
Jskid
don't understand part of answer to contrapositive question
Find the contrapositive. If there exists real numbers x and y with x $\displaystyle \neq$ y and $\displaystyle x^{2}+x y+y^{2}+x+y=0$ then f is not one-to-one

I got:
if f is one-to-one then for all real numbers x and y with x=y or $\displaystyle x^{2}+x y+y^{2}+x+y \neq 0$
Why is it still = and not $\displaystyle \neq$?
• Sep 8th 2011, 09:26 PM
pickslides
Re: don't understand part of answer to contrapositive question
Hey Jskid

When thinking about the contrapositive consider this.

The original statement: if m then n

Then the contrapositive will be: if not n then not m.

does this help?

P.S. to avoid those nasty latex errors use "tex" tags not "math" tags.
• Sep 8th 2011, 10:09 PM
Jskid
Re: don't understand part of answer to contrapositive question
Quote:

Originally Posted by pickslides
Hey Jskid

When thinking about the contrapositive consider this.

The original statement: if m then n

Then the contrapositive will be: if not n then not m.

does this help?

P.S. to avoid those nasty latex errors use "tex" tags not "math" tags.

A little.
So n is "If there exists real numbers x and y with $\displaystyle x \neq y$ and $\displaystyle x^{2}+x y+y^{2}+x+y=0$" right?
Why does the negation change x=y to $\displaystyle x \neq y$ the "and" to "or" but not $\displaystyle x^{2}+x y+y^{2}+x+y = 0$ to $\displaystyle x^{2}+x y+y^{2}+x+y \neq 0$?
• Sep 9th 2011, 02:51 AM
emakarov
Re: don't understand part of answer to contrapositive question
We have $\displaystyle \neg(A\land B)\Leftrightarrow\neg A\lor\neg B\Leftrightarrow A\to\neg B$.

Quote:

Originally Posted by Jskid
Find the contrapositive. If there exists real numbers x and y with x $\displaystyle \neq$ y and $\displaystyle x^{2}+x y+y^{2}+x+y=0$ then f is not one-to-one

I got:
if f is one-to-one then for all real numbers x and y with x=y or $\displaystyle x^{2}+x y+y^{2}+x+y \neq 0$

Should be: "...then for all real numbers x and y, it is the case that x=y or $\displaystyle x^{2}+x y+y^{2}+x+y \neq 0$." (The phrase "it is the case that" is inserted just to avoid two mathematical expressions with no words between them.) Alternatively, "...then for all real x and y, if $\displaystyle x\ne y$, then $\displaystyle x^{2}+x y+y^{2}+x+y \neq 0$."

Quote:

Originally Posted by Jskid
Why is it still = and not $\displaystyle \neq$?

What is "still" = ?. The assumption was $\displaystyle x\ne y\land x^{2}+x y+y^{2}+x+y=0$; its negation is $\displaystyle x=y\lor x^{2}+x y+y^{2}+x+y\ne0$.

Quote:

Originally Posted by Jskid
So n is "If there exists real numbers x and y with $\displaystyle x \neq y$ and $\displaystyle x^{2}+x y+y^{2}+x+y=0$" right?

Yes, but without the initial "If."

Quote:

Originally Posted by Jskid
Why does the negation change x=y to $\displaystyle x \neq y$ the "and" to "or" but not $\displaystyle x^{2}+x y+y^{2}+x+y = 0$ to $\displaystyle x^{2}+x y+y^{2}+x+y \neq 0$?

The last equation does change to inequation.
• Sep 11th 2011, 11:52 PM
Jskid
Re: don't understand part of answer to contrapositive question
The answer key has
If f is one-to-one, then for all real numbers x and y either x=y or $\displaystyle x^2+xy+y^2+x+y=0$

is the end a typo? Shouldn't it be $\displaystyle \neq 0$?
• Sep 12th 2011, 01:27 AM
emakarov
Re: don't understand part of answer to contrapositive question
Yes, I think so.