Prove that $f(x) = x ^ 2n+1$ is injective.

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1
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So I need to prove that

$$f : Bbb R rightarrow Bbb R,quad f(x) = x ^2n+1,$$

where $ninBbb N$ is an injective function. Or rather I need to prove that the function of a number to an odd exponent is injective.

I've been trying to prove it with a contradiction, assuming that there are 2 different values of $x$ for the same $y$ but then I get stuck.

Any help would be highly appreciated!

proof-explanation

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edited Jul 17 at 12:01

Shaun

7,41492972

asked Jul 17 at 11:58

Relikus

141

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  It's increasing.
  – Lord Shark the Unknown
  Jul 17 at 11:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Here's a MathJax tutorial :)
  – Shaun
  Jul 17 at 12:00
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

So I need to prove that

$$f : Bbb R rightarrow Bbb R,quad f(x) = x ^2n+1,$$

where $ninBbb N$ is an injective function. Or rather I need to prove that the function of a number to an odd exponent is injective.

I've been trying to prove it with a contradiction, assuming that there are 2 different values of $x$ for the same $y$ but then I get stuck.

Any help would be highly appreciated!

proof-explanation

share|cite|improve this question

edited Jul 17 at 12:01

Shaun

7,41492972

asked Jul 17 at 11:58

Relikus

141

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  It's increasing.
  – Lord Shark the Unknown
  Jul 17 at 11:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Here's a MathJax tutorial :)
  – Shaun
  Jul 17 at 12:00
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

So I need to prove that

$$f : Bbb R rightarrow Bbb R,quad f(x) = x ^2n+1,$$

where $ninBbb N$ is an injective function. Or rather I need to prove that the function of a number to an odd exponent is injective.

I've been trying to prove it with a contradiction, assuming that there are 2 different values of $x$ for the same $y$ but then I get stuck.

Any help would be highly appreciated!

proof-explanation

share|cite|improve this question

edited Jul 17 at 12:01

Shaun

7,41492972

asked Jul 17 at 11:58

Relikus

141

So I need to prove that

$$f : Bbb R rightarrow Bbb R,quad f(x) = x ^2n+1,$$

where $ninBbb N$ is an injective function. Or rather I need to prove that the function of a number to an odd exponent is injective.

I've been trying to prove it with a contradiction, assuming that there are 2 different values of $x$ for the same $y$ but then I get stuck.

Any help would be highly appreciated!

proof-explanation

share|cite|improve this question

edited Jul 17 at 12:01

Shaun

7,41492972

asked Jul 17 at 11:58

Relikus

141

share|cite|improve this question

share|cite|improve this question

edited Jul 17 at 12:01

Shaun

7,41492972

edited Jul 17 at 12:01

Shaun

7,41492972

edited Jul 17 at 12:01

Shaun

7,41492972

7,41492972

asked Jul 17 at 11:58

Relikus

141

asked Jul 17 at 11:58

Relikus

141

asked Jul 17 at 11:58

Relikus

141

141

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  It's increasing.
  – Lord Shark the Unknown
  Jul 17 at 11:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Here's a MathJax tutorial :)
  – Shaun
  Jul 17 at 12:00
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  It's increasing.
  – Lord Shark the Unknown
  Jul 17 at 11:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Here's a MathJax tutorial :)
  – Shaun
  Jul 17 at 12:00
  

  
  
  
  

3

3

It's increasing.
– Lord Shark the Unknown
Jul 17 at 11:59

It's increasing.
– Lord Shark the Unknown
Jul 17 at 11:59

Here's a MathJax tutorial :)
– Shaun
Jul 17 at 12:00

Here's a MathJax tutorial :)
– Shaun
Jul 17 at 12:00

add a comment | 

2 Answers
2

active

oldest

votes

up vote
7
down vote

Take $x, y$ such that $x^2n+1 = y^2n+1$. If $x = 0$, the expression becomes just $0 = y^2n+1$, which means that $y = 0$ as well. Same for $y = 0$ implying $x = 0$.

Assuming $x, yneq 0$, they must clearly have the same sign, because otherwise $x^2n+1$ and $y^2n+1$ wouldn't have the same sign, and couldn't possibly be equal. Finally, note that $x^2n+1 = y^2n+1$ may be rewritten to $x^2n+1 - y^2n+1 = 0$, which gives
$$
0 = x^2n+1-y^2n+1 = (x-y)(x^2n + x^2n-1y + x^2n-2y^2 + cdots + xy^2n-1 + y^2n)
$$
Because $x$ and $y$ are both non-zero, and have the same sign, the second factor is clearly positive (it's a sum of $2n+1$ strictly positive terms). So if the product is $0$, that means that the first factor must be $0$. This gives $x-y = 0$ or $x = y$.

share|cite|improve this answer

edited Jul 17 at 14:00

answered Jul 17 at 12:13

Arthur

98.9k793175

add a comment | 

up vote
1
down vote

Suppose $x^2n+1 = y^2n+1$. The numbers $x,y$ must have same sign. We're obviously interested in the nonzero case so $$fracxy = left (fracyxright )^2n $$
If $x>y$, then $1<fracxy = left (fracyxright )^2n<1$?!
Similarly, if $x<y$, then $1>fracxy = left (fracyxright )^2n>1$?!

---

Arthur's much better explanation: for the nonzero case
$$1 = left (fracxyright )^2n+1 implies 1 = fracxy $$
since the number $1$ has exactly one real $2n+1$st root.

share|cite|improve this answer

edited Jul 17 at 12:45

answered Jul 17 at 12:21

Alvin Lepik

2,053819

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Somewhat easier to follow: $1 = fracy^2n+1x^2n+1 = left(frac yxright)^2n+1$, giving $frac yx = 1$ since $a^2n+1 = 1$ only has one real solution.
  – Arthur
  Jul 17 at 12:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arthur , it's summer, I'm rusty :(
  – Alvin Lepik
  Jul 17 at 12:43
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
7
down vote

Take $x, y$ such that $x^2n+1 = y^2n+1$. If $x = 0$, the expression becomes just $0 = y^2n+1$, which means that $y = 0$ as well. Same for $y = 0$ implying $x = 0$.

Assuming $x, yneq 0$, they must clearly have the same sign, because otherwise $x^2n+1$ and $y^2n+1$ wouldn't have the same sign, and couldn't possibly be equal. Finally, note that $x^2n+1 = y^2n+1$ may be rewritten to $x^2n+1 - y^2n+1 = 0$, which gives
$$
0 = x^2n+1-y^2n+1 = (x-y)(x^2n + x^2n-1y + x^2n-2y^2 + cdots + xy^2n-1 + y^2n)
$$
Because $x$ and $y$ are both non-zero, and have the same sign, the second factor is clearly positive (it's a sum of $2n+1$ strictly positive terms). So if the product is $0$, that means that the first factor must be $0$. This gives $x-y = 0$ or $x = y$.

share|cite|improve this answer

edited Jul 17 at 14:00

answered Jul 17 at 12:13

Arthur

98.9k793175

add a comment | 

up vote
7
down vote

Take $x, y$ such that $x^2n+1 = y^2n+1$. If $x = 0$, the expression becomes just $0 = y^2n+1$, which means that $y = 0$ as well. Same for $y = 0$ implying $x = 0$.

Assuming $x, yneq 0$, they must clearly have the same sign, because otherwise $x^2n+1$ and $y^2n+1$ wouldn't have the same sign, and couldn't possibly be equal. Finally, note that $x^2n+1 = y^2n+1$ may be rewritten to $x^2n+1 - y^2n+1 = 0$, which gives
$$
0 = x^2n+1-y^2n+1 = (x-y)(x^2n + x^2n-1y + x^2n-2y^2 + cdots + xy^2n-1 + y^2n)
$$
Because $x$ and $y$ are both non-zero, and have the same sign, the second factor is clearly positive (it's a sum of $2n+1$ strictly positive terms). So if the product is $0$, that means that the first factor must be $0$. This gives $x-y = 0$ or $x = y$.

share|cite|improve this answer

edited Jul 17 at 14:00

answered Jul 17 at 12:13

Arthur

98.9k793175

add a comment | 

up vote
7
down vote

up vote
7
down vote

Take $x, y$ such that $x^2n+1 = y^2n+1$. If $x = 0$, the expression becomes just $0 = y^2n+1$, which means that $y = 0$ as well. Same for $y = 0$ implying $x = 0$.

Assuming $x, yneq 0$, they must clearly have the same sign, because otherwise $x^2n+1$ and $y^2n+1$ wouldn't have the same sign, and couldn't possibly be equal. Finally, note that $x^2n+1 = y^2n+1$ may be rewritten to $x^2n+1 - y^2n+1 = 0$, which gives
$$
0 = x^2n+1-y^2n+1 = (x-y)(x^2n + x^2n-1y + x^2n-2y^2 + cdots + xy^2n-1 + y^2n)
$$
Because $x$ and $y$ are both non-zero, and have the same sign, the second factor is clearly positive (it's a sum of $2n+1$ strictly positive terms). So if the product is $0$, that means that the first factor must be $0$. This gives $x-y = 0$ or $x = y$.

share|cite|improve this answer

edited Jul 17 at 14:00

answered Jul 17 at 12:13

Arthur

98.9k793175

Take $x, y$ such that $x^2n+1 = y^2n+1$. If $x = 0$, the expression becomes just $0 = y^2n+1$, which means that $y = 0$ as well. Same for $y = 0$ implying $x = 0$.

Assuming $x, yneq 0$, they must clearly have the same sign, because otherwise $x^2n+1$ and $y^2n+1$ wouldn't have the same sign, and couldn't possibly be equal. Finally, note that $x^2n+1 = y^2n+1$ may be rewritten to $x^2n+1 - y^2n+1 = 0$, which gives
$$
0 = x^2n+1-y^2n+1 = (x-y)(x^2n + x^2n-1y + x^2n-2y^2 + cdots + xy^2n-1 + y^2n)
$$
Because $x$ and $y$ are both non-zero, and have the same sign, the second factor is clearly positive (it's a sum of $2n+1$ strictly positive terms). So if the product is $0$, that means that the first factor must be $0$. This gives $x-y = 0$ or $x = y$.

share|cite|improve this answer

edited Jul 17 at 14:00

answered Jul 17 at 12:13

Arthur

98.9k793175

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 14:00

edited Jul 17 at 14:00

edited Jul 17 at 14:00

answered Jul 17 at 12:13

Arthur

98.9k793175

answered Jul 17 at 12:13

Arthur

98.9k793175

answered Jul 17 at 12:13

Arthur

98.9k793175

98.9k793175

add a comment | 

add a comment | 

up vote
1
down vote

Suppose $x^2n+1 = y^2n+1$. The numbers $x,y$ must have same sign. We're obviously interested in the nonzero case so $$fracxy = left (fracyxright )^2n $$
If $x>y$, then $1<fracxy = left (fracyxright )^2n<1$?!
Similarly, if $x<y$, then $1>fracxy = left (fracyxright )^2n>1$?!

---

Arthur's much better explanation: for the nonzero case
$$1 = left (fracxyright )^2n+1 implies 1 = fracxy $$
since the number $1$ has exactly one real $2n+1$st root.

share|cite|improve this answer

edited Jul 17 at 12:45

answered Jul 17 at 12:21

Alvin Lepik

2,053819

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Somewhat easier to follow: $1 = fracy^2n+1x^2n+1 = left(frac yxright)^2n+1$, giving $frac yx = 1$ since $a^2n+1 = 1$ only has one real solution.
  – Arthur
  Jul 17 at 12:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arthur , it's summer, I'm rusty :(
  – Alvin Lepik
  Jul 17 at 12:43
  

  
  
  
  

add a comment | 

up vote
1
down vote

Suppose $x^2n+1 = y^2n+1$. The numbers $x,y$ must have same sign. We're obviously interested in the nonzero case so $$fracxy = left (fracyxright )^2n $$
If $x>y$, then $1<fracxy = left (fracyxright )^2n<1$?!
Similarly, if $x<y$, then $1>fracxy = left (fracyxright )^2n>1$?!

---

Arthur's much better explanation: for the nonzero case
$$1 = left (fracxyright )^2n+1 implies 1 = fracxy $$
since the number $1$ has exactly one real $2n+1$st root.

share|cite|improve this answer

edited Jul 17 at 12:45

answered Jul 17 at 12:21

Alvin Lepik

2,053819

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Somewhat easier to follow: $1 = fracy^2n+1x^2n+1 = left(frac yxright)^2n+1$, giving $frac yx = 1$ since $a^2n+1 = 1$ only has one real solution.
  – Arthur
  Jul 17 at 12:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arthur , it's summer, I'm rusty :(
  – Alvin Lepik
  Jul 17 at 12:43
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Suppose $x^2n+1 = y^2n+1$. The numbers $x,y$ must have same sign. We're obviously interested in the nonzero case so $$fracxy = left (fracyxright )^2n $$
If $x>y$, then $1<fracxy = left (fracyxright )^2n<1$?!
Similarly, if $x<y$, then $1>fracxy = left (fracyxright )^2n>1$?!

---

Arthur's much better explanation: for the nonzero case
$$1 = left (fracxyright )^2n+1 implies 1 = fracxy $$
since the number $1$ has exactly one real $2n+1$st root.

share|cite|improve this answer

edited Jul 17 at 12:45

answered Jul 17 at 12:21

Alvin Lepik

2,053819

Suppose $x^2n+1 = y^2n+1$. The numbers $x,y$ must have same sign. We're obviously interested in the nonzero case so $$fracxy = left (fracyxright )^2n $$
If $x>y$, then $1<fracxy = left (fracyxright )^2n<1$?!
Similarly, if $x<y$, then $1>fracxy = left (fracyxright )^2n>1$?!

---

Arthur's much better explanation: for the nonzero case
$$1 = left (fracxyright )^2n+1 implies 1 = fracxy $$
since the number $1$ has exactly one real $2n+1$st root.

share|cite|improve this answer

edited Jul 17 at 12:45

answered Jul 17 at 12:21

Alvin Lepik

2,053819

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 12:45

edited Jul 17 at 12:45

edited Jul 17 at 12:45

answered Jul 17 at 12:21

Alvin Lepik

2,053819

answered Jul 17 at 12:21

Alvin Lepik

2,053819

answered Jul 17 at 12:21

Alvin Lepik

2,053819

2,053819

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Somewhat easier to follow: $1 = fracy^2n+1x^2n+1 = left(frac yxright)^2n+1$, giving $frac yx = 1$ since $a^2n+1 = 1$ only has one real solution.
  – Arthur
  Jul 17 at 12:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arthur , it's summer, I'm rusty :(
  – Alvin Lepik
  Jul 17 at 12:43
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Somewhat easier to follow: $1 = fracy^2n+1x^2n+1 = left(frac yxright)^2n+1$, giving $frac yx = 1$ since $a^2n+1 = 1$ only has one real solution.
  – Arthur
  Jul 17 at 12:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arthur , it's summer, I'm rusty :(
  – Alvin Lepik
  Jul 17 at 12:43
  

  
  
  
  

Somewhat easier to follow: $1 = fracy^2n+1x^2n+1 = left(frac yxright)^2n+1$, giving $frac yx = 1$ since $a^2n+1 = 1$ only has one real solution.
– Arthur
Jul 17 at 12:31

Somewhat easier to follow: $1 = fracy^2n+1x^2n+1 = left(frac yxright)^2n+1$, giving $frac yx = 1$ since $a^2n+1 = 1$ only has one real solution.
– Arthur
Jul 17 at 12:31

@Arthur , it's summer, I'm rusty :(
– Alvin Lepik
Jul 17 at 12:43

@Arthur , it's summer, I'm rusty :(
– Alvin Lepik
Jul 17 at 12:43

add a comment | 

 

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