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Show the function $f(n)=dfrac (-1)^n (2n-1)+1 4$ is a surjection

Why isn't this function $f:mathbb N to mathcal P(mathbb N)$ a surjection?How does a special case prove a surjection?Which function is an injection but NOT A SURJECTIONHelp prove $f:X rightarrow Y$ is an injection $Leftrightarrow$ $f:Xrightarrow Y$ is a surjection when $|X|=|Y|$Cantor's Theorem (surjection vs bijection)Surjection $f$ induces surjection $mathcal P (f)$ on power setsHow do I show this is a surjection?Find an explicit surjection $f : mathbbZ_+ to mathbbZ_+ times mathbbZ_+$prove that the function is surjective but not injective.Show that the following function is continuous

1

1

$begingroup$

• Let $f:mathbbNrightarrowmathbbZ$ be a function which $f(n)=dfrac (-1)^n (2n-1)+1 4$. Show that $f$ is surjection.

Proof. Let $minmathbbZ$. We need to find $ninmathbbN$ such that $f(n)=m$.Let $m=dfrac (-1)^n (2n-1)+1 4$ for $ninmathbbN.$ If $n$ is even then we get

$$m=dfrac (-1)^n (2n-1)+1 4,$$

$$4m=2n,$$

$$2m=ninmathbbN.$$

If $n$ is odd, then we get

$$4m=-2n+1+1$$

$$1-2m=ninmathbbN.$$

Therefore, can we say $f$ is surjective?

functions elementary-set-theory

share|cite|improve this question

edited Apr 21 at 15:47

YuiTo Cheng

2,90141139

asked Apr 21 at 14:38

PozcuKushimotoStreetPozcuKushimotoStreet

1,425923

$endgroup$

add a comment | 

1

1

$begingroup$

• Let $f:mathbbNrightarrowmathbbZ$ be a function which $f(n)=dfrac (-1)^n (2n-1)+1 4$. Show that $f$ is surjection.

Proof. Let $minmathbbZ$. We need to find $ninmathbbN$ such that $f(n)=m$.Let $m=dfrac (-1)^n (2n-1)+1 4$ for $ninmathbbN.$ If $n$ is even then we get

$$m=dfrac (-1)^n (2n-1)+1 4,$$

$$4m=2n,$$

$$2m=ninmathbbN.$$

If $n$ is odd, then we get

$$4m=-2n+1+1$$

$$1-2m=ninmathbbN.$$

Therefore, can we say $f$ is surjective?

functions elementary-set-theory

share|cite|improve this question

edited Apr 21 at 15:47

YuiTo Cheng

2,90141139

asked Apr 21 at 14:38

PozcuKushimotoStreetPozcuKushimotoStreet

1,425923

$endgroup$

add a comment | 

$begingroup$

• Let $f:mathbbNrightarrowmathbbZ$ be a function which $f(n)=dfrac (-1)^n (2n-1)+1 4$. Show that $f$ is surjection.

Proof. Let $minmathbbZ$. We need to find $ninmathbbN$ such that $f(n)=m$.Let $m=dfrac (-1)^n (2n-1)+1 4$ for $ninmathbbN.$ If $n$ is even then we get

$$m=dfrac (-1)^n (2n-1)+1 4,$$

$$4m=2n,$$

$$2m=ninmathbbN.$$

If $n$ is odd, then we get

$$4m=-2n+1+1$$

$$1-2m=ninmathbbN.$$

Therefore, can we say $f$ is surjective?

functions elementary-set-theory

share|cite|improve this question

edited Apr 21 at 15:47

YuiTo Cheng

2,90141139

asked Apr 21 at 14:38

PozcuKushimotoStreetPozcuKushimotoStreet

1,425923

$endgroup$

share|cite|improve this question

edited Apr 21 at 15:47

YuiTo Cheng

2,90141139

asked Apr 21 at 14:38

PozcuKushimotoStreetPozcuKushimotoStreet

1,425923

share|cite|improve this question

edited Apr 21 at 15:47

YuiTo Cheng

2,90141139

asked Apr 21 at 14:38

PozcuKushimotoStreetPozcuKushimotoStreet

1,425923

edited Apr 21 at 15:47

YuiTo Cheng

2,90141139

edited Apr 21 at 15:47

YuiTo Cheng

2,90141139

asked Apr 21 at 14:38

PozcuKushimotoStreetPozcuKushimotoStreet

1,425923

asked Apr 21 at 14:38

PozcuKushimotoStreetPozcuKushimotoStreet

1,425923

3 Answers
3

active

oldest

votes

2

$begingroup$

You can leave it more explicit

Let $n in mathbbN$

If $n$ is even, $n = 2k$ for some $k in mathbbN$

Then $f(2k) = f(n) =frac(-1)^n(2n-1)+14 = frac(2n-1)+14 = frac2n4 = frac4k4 = k $

So, for each $m in mathbbN$, if we take $n = 2m$, we have that $f(n) = f(2m) = m$

Now if $n$ is odd, we have $n = 2k -1$ for some $k in mathbbN$

Then $f(2k-1) = f(n) = frac(-1)^n(2n-1)+14 = frac(-2n +1) +14 = frac(-2n + 2)4 = frac-2(2k-1) +24 = frac-4k + 44 = -k + 1$

So if $m in mathbbZ, m ≤ 0$ then $m - 1≤-1$ so $-m +1≥ 1$
Then $-m + 1 in mathbbN$
And if we take $n = 2(-m+1) -1$ then we have $f(n) = f(2(-m+1)-1) = -(-m+1) + 1 = m$

Then $f$ is surjrctive.

share|cite|improve this answer

answered Apr 21 at 15:11

ZAFZAF

5387

$endgroup$

add a comment | 

3

$begingroup$

Your proof is a little backward. You need to find $n in mathbbN$ such that $f(n)=m$. So you can't say "let $m= f(n)$", because you don't know that there is such $n$ a priori. As such, it doesn't make sense to say "if $n$ is even" or "if $n$ is odd", since you have not demonstrated that $n$ exists. What you have written down is like the scratch work to figure out how the proof should go, but it does not constitute a proof.

Notice that:
beginalign*
f(1)&=0\
f(2) &= 1 \
f(3) &= -1\
f(4) &=2\
f(5)&=-2 \
&vdots
endalign*

Now consider separately the cases $m=0$, $m>0$, and $m<0$. In each case, come up with a formula for $n$ in terms of $m$ that will satisfy $f(n)=m$.

share|cite|improve this answer

answered Apr 21 at 14:51

kccukccu

12k11231

$endgroup$

add a comment | 

2

$begingroup$

a little algebra shows:
$$
f(n+2) - f(n) = (-1)^n
$$

since $f(0) = 0$, then by a telescoping sum:
$$
f(0+2m) = sum_k=1^m (-1)^0 = m
$$

similarly, $f(1) = 0$, and so
$$
f(1+2m) = sum_k=1^m (-1)^1 = -m
$$

share|cite|improve this answer

answered Apr 21 at 15:31

David HoldenDavid Holden

15k21226

$endgroup$

add a comment | 

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2

$begingroup$

You can leave it more explicit

Let $n in mathbbN$

If $n$ is even, $n = 2k$ for some $k in mathbbN$

Then $f(2k) = f(n) =frac(-1)^n(2n-1)+14 = frac(2n-1)+14 = frac2n4 = frac4k4 = k $

So, for each $m in mathbbN$, if we take $n = 2m$, we have that $f(n) = f(2m) = m$

Now if $n$ is odd, we have $n = 2k -1$ for some $k in mathbbN$

Then $f(2k-1) = f(n) = frac(-1)^n(2n-1)+14 = frac(-2n +1) +14 = frac(-2n + 2)4 = frac-2(2k-1) +24 = frac-4k + 44 = -k + 1$

So if $m in mathbbZ, m ≤ 0$ then $m - 1≤-1$ so $-m +1≥ 1$
Then $-m + 1 in mathbbN$
And if we take $n = 2(-m+1) -1$ then we have $f(n) = f(2(-m+1)-1) = -(-m+1) + 1 = m$

Then $f$ is surjrctive.

share|cite|improve this answer

answered Apr 21 at 15:11

ZAFZAF

5387

$endgroup$

add a comment | 

2

$begingroup$

You can leave it more explicit

Let $n in mathbbN$

If $n$ is even, $n = 2k$ for some $k in mathbbN$

Then $f(2k) = f(n) =frac(-1)^n(2n-1)+14 = frac(2n-1)+14 = frac2n4 = frac4k4 = k $

So, for each $m in mathbbN$, if we take $n = 2m$, we have that $f(n) = f(2m) = m$

Now if $n$ is odd, we have $n = 2k -1$ for some $k in mathbbN$

Then $f(2k-1) = f(n) = frac(-1)^n(2n-1)+14 = frac(-2n +1) +14 = frac(-2n + 2)4 = frac-2(2k-1) +24 = frac-4k + 44 = -k + 1$

So if $m in mathbbZ, m ≤ 0$ then $m - 1≤-1$ so $-m +1≥ 1$
Then $-m + 1 in mathbbN$
And if we take $n = 2(-m+1) -1$ then we have $f(n) = f(2(-m+1)-1) = -(-m+1) + 1 = m$

Then $f$ is surjrctive.

share|cite|improve this answer

answered Apr 21 at 15:11

ZAFZAF

5387

$endgroup$

add a comment | 

$begingroup$

You can leave it more explicit

Let $n in mathbbN$

If $n$ is even, $n = 2k$ for some $k in mathbbN$

Then $f(2k) = f(n) =frac(-1)^n(2n-1)+14 = frac(2n-1)+14 = frac2n4 = frac4k4 = k $

So, for each $m in mathbbN$, if we take $n = 2m$, we have that $f(n) = f(2m) = m$

Now if $n$ is odd, we have $n = 2k -1$ for some $k in mathbbN$

Then $f(2k-1) = f(n) = frac(-1)^n(2n-1)+14 = frac(-2n +1) +14 = frac(-2n + 2)4 = frac-2(2k-1) +24 = frac-4k + 44 = -k + 1$

So if $m in mathbbZ, m ≤ 0$ then $m - 1≤-1$ so $-m +1≥ 1$
Then $-m + 1 in mathbbN$
And if we take $n = 2(-m+1) -1$ then we have $f(n) = f(2(-m+1)-1) = -(-m+1) + 1 = m$

Then $f$ is surjrctive.

share|cite|improve this answer

answered Apr 21 at 15:11

ZAFZAF

5387

$endgroup$

share|cite|improve this answer

answered Apr 21 at 15:11

ZAFZAF

5387

answered Apr 21 at 15:11

ZAFZAF

5387

answered Apr 21 at 15:11

ZAFZAF

5387

3

$begingroup$

Your proof is a little backward. You need to find $n in mathbbN$ such that $f(n)=m$. So you can't say "let $m= f(n)$", because you don't know that there is such $n$ a priori. As such, it doesn't make sense to say "if $n$ is even" or "if $n$ is odd", since you have not demonstrated that $n$ exists. What you have written down is like the scratch work to figure out how the proof should go, but it does not constitute a proof.

Notice that:
beginalign*
f(1)&=0\
f(2) &= 1 \
f(3) &= -1\
f(4) &=2\
f(5)&=-2 \
&vdots
endalign*

Now consider separately the cases $m=0$, $m>0$, and $m<0$. In each case, come up with a formula for $n$ in terms of $m$ that will satisfy $f(n)=m$.

share|cite|improve this answer

answered Apr 21 at 14:51

kccukccu

12k11231

$endgroup$

add a comment | 

3

$begingroup$

Your proof is a little backward. You need to find $n in mathbbN$ such that $f(n)=m$. So you can't say "let $m= f(n)$", because you don't know that there is such $n$ a priori. As such, it doesn't make sense to say "if $n$ is even" or "if $n$ is odd", since you have not demonstrated that $n$ exists. What you have written down is like the scratch work to figure out how the proof should go, but it does not constitute a proof.

Notice that:
beginalign*
f(1)&=0\
f(2) &= 1 \
f(3) &= -1\
f(4) &=2\
f(5)&=-2 \
&vdots
endalign*

Now consider separately the cases $m=0$, $m>0$, and $m<0$. In each case, come up with a formula for $n$ in terms of $m$ that will satisfy $f(n)=m$.

share|cite|improve this answer

answered Apr 21 at 14:51

kccukccu

12k11231

$endgroup$

add a comment | 

$begingroup$

Your proof is a little backward. You need to find $n in mathbbN$ such that $f(n)=m$. So you can't say "let $m= f(n)$", because you don't know that there is such $n$ a priori. As such, it doesn't make sense to say "if $n$ is even" or "if $n$ is odd", since you have not demonstrated that $n$ exists. What you have written down is like the scratch work to figure out how the proof should go, but it does not constitute a proof.

Notice that:
beginalign*
f(1)&=0\
f(2) &= 1 \
f(3) &= -1\
f(4) &=2\
f(5)&=-2 \
&vdots
endalign*

Now consider separately the cases $m=0$, $m>0$, and $m<0$. In each case, come up with a formula for $n$ in terms of $m$ that will satisfy $f(n)=m$.

share|cite|improve this answer

answered Apr 21 at 14:51

kccukccu

12k11231

$endgroup$

share|cite|improve this answer

answered Apr 21 at 14:51

kccukccu

12k11231

answered Apr 21 at 14:51

kccukccu

12k11231

answered Apr 21 at 14:51

kccukccu

12k11231

2

$begingroup$

a little algebra shows:
$$
f(n+2) - f(n) = (-1)^n
$$

since $f(0) = 0$, then by a telescoping sum:
$$
f(0+2m) = sum_k=1^m (-1)^0 = m
$$

similarly, $f(1) = 0$, and so
$$
f(1+2m) = sum_k=1^m (-1)^1 = -m
$$

share|cite|improve this answer

answered Apr 21 at 15:31

David HoldenDavid Holden

15k21226

$endgroup$

add a comment | 

2

$begingroup$

a little algebra shows:
$$
f(n+2) - f(n) = (-1)^n
$$

since $f(0) = 0$, then by a telescoping sum:
$$
f(0+2m) = sum_k=1^m (-1)^0 = m
$$

similarly, $f(1) = 0$, and so
$$
f(1+2m) = sum_k=1^m (-1)^1 = -m
$$

share|cite|improve this answer

answered Apr 21 at 15:31

David HoldenDavid Holden

15k21226

$endgroup$

add a comment | 

$begingroup$

a little algebra shows:
$$
f(n+2) - f(n) = (-1)^n
$$

since $f(0) = 0$, then by a telescoping sum:
$$
f(0+2m) = sum_k=1^m (-1)^0 = m
$$

similarly, $f(1) = 0$, and so
$$
f(1+2m) = sum_k=1^m (-1)^1 = -m
$$

share|cite|improve this answer

answered Apr 21 at 15:31

David HoldenDavid Holden

15k21226

$endgroup$

share|cite|improve this answer

answered Apr 21 at 15:31

David HoldenDavid Holden

15k21226

answered Apr 21 at 15:31

David HoldenDavid Holden

15k21226

answered Apr 21 at 15:31

David HoldenDavid Holden

15k21226