Proof that function is bijective: f(x) = x+1 (x odd), x - 1 (x even)

Adesh

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Proof that function is bijective.

f is R to R.
f(x)={x+1, if x is odd , x-1, if x is even}
Please answer stepwise because I'm suffering glitch in proving one-one.
 
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Prove that the following function is bijective:

. . . . .\(\displaystyle f(x)\, =\, \begin{cases}x\, +\, 1,\, &\mbox{ if }\, x\, \mbox{is odd} \\ x\, -\, 1,\, &\mbox{ if }\, x\, \mbox{ is even}\end{cases}\)

Please provide stepwise answer.
 
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Prove that the following function is bijective:

. . . . .\(\displaystyle f(x)\, =\, \begin{cases}x\, +\, 1,\, &\mbox{ if }\, x\, \mbox{is odd} \\ x\, -\, 1,\, &\mbox{ if }\, x\, \mbox{ is even}\end{cases}\)

Please provide stepwise answer.
I'm sorry, but the policy you saw in the "Read Before Posting" announcement remains in effect; namely, we don't "do" students' work for them, nor do we provide complete solutions.

Kindly please reply with a clear listing of your thoughts and efforts so far (starting with your book's definition of a "bijective" function), so we can see where things are bogging down. Thank you! ;)
 
f:R--R
f(x)={x+1, if x is odd , x-1, if x is even}
ALSO please correct the notation. It cannot be that.

That is a function from reals to reals. Real numbers are not even or odd.
Do you mean integers?
 
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