Properties of the composition law $a @ b= ab + b + a$ on the real numbers [closed]

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

The operation $@$ is defined on the real numbers as $a @ b= ab + b + a$

a) Show that $0$ is an identity for the operation.

b) Show that some real numbers have inverses under the operation.

c) Find a counter-example to show that, for this operation inverses do not exist for all the real numbers.

I have so far came to understand as far as: $a, b,c in mathbbR$
Identity: Ia= aI= a
@ b I= @ b + b + I
I @ b = I b + b + I

group-theory

share|cite|improve this question

edited Jul 23 at 11:02

Did

242k23208443

asked Jul 23 at 10:10

shayy

62

closed as off-topic by John Ma, Mathmo123, Taroccoesbrocco, Arnaud D., Asaf Karagila Aug 1 at 12:48

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Mathmo123, Taroccoesbrocco

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, just write it out. What is $0@1$, for example? $0@7$? $0@b$?
  – lulu
  Jul 23 at 10:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  To prove $0$ is the identity, show that $a@0=0@a=a$ for an arbitrary choice of $a$
  – aidangallagher4
  Jul 23 at 10:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  To understand (and prove) in one strike all these properties (and how your TA built their exercise), note that $@$ is really the ordinary product in disguise, since $$(a@b)+1=(a+1)cdot(b+1)$$
  – Did
  Jul 23 at 11:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Possible duplicate of Proving $G$ is a group under a specific operation
  – Arnaud D.
  Aug 1 at 12:47
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

The operation $@$ is defined on the real numbers as $a @ b= ab + b + a$

a) Show that $0$ is an identity for the operation.

b) Show that some real numbers have inverses under the operation.

c) Find a counter-example to show that, for this operation inverses do not exist for all the real numbers.

I have so far came to understand as far as: $a, b,c in mathbbR$
Identity: Ia= aI= a
@ b I= @ b + b + I
I @ b = I b + b + I

group-theory

share|cite|improve this question

edited Jul 23 at 11:02

Did

242k23208443

asked Jul 23 at 10:10

shayy

62

closed as off-topic by John Ma, Mathmo123, Taroccoesbrocco, Arnaud D., Asaf Karagila Aug 1 at 12:48

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Mathmo123, Taroccoesbrocco

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, just write it out. What is $0@1$, for example? $0@7$? $0@b$?
  – lulu
  Jul 23 at 10:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  To prove $0$ is the identity, show that $a@0=0@a=a$ for an arbitrary choice of $a$
  – aidangallagher4
  Jul 23 at 10:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  To understand (and prove) in one strike all these properties (and how your TA built their exercise), note that $@$ is really the ordinary product in disguise, since $$(a@b)+1=(a+1)cdot(b+1)$$
  – Did
  Jul 23 at 11:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Possible duplicate of Proving $G$ is a group under a specific operation
  – Arnaud D.
  Aug 1 at 12:47
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

The operation $@$ is defined on the real numbers as $a @ b= ab + b + a$

a) Show that $0$ is an identity for the operation.

b) Show that some real numbers have inverses under the operation.

c) Find a counter-example to show that, for this operation inverses do not exist for all the real numbers.

I have so far came to understand as far as: $a, b,c in mathbbR$
Identity: Ia= aI= a
@ b I= @ b + b + I
I @ b = I b + b + I

group-theory

share|cite|improve this question

edited Jul 23 at 11:02

Did

242k23208443

asked Jul 23 at 10:10

shayy

62

The operation $@$ is defined on the real numbers as $a @ b= ab + b + a$

a) Show that $0$ is an identity for the operation.

b) Show that some real numbers have inverses under the operation.

c) Find a counter-example to show that, for this operation inverses do not exist for all the real numbers.

I have so far came to understand as far as: $a, b,c in mathbbR$
Identity: Ia= aI= a
@ b I= @ b + b + I
I @ b = I b + b + I

group-theory

share|cite|improve this question

edited Jul 23 at 11:02

Did

242k23208443

asked Jul 23 at 10:10

shayy

62

share|cite|improve this question

share|cite|improve this question

edited Jul 23 at 11:02

Did

242k23208443

edited Jul 23 at 11:02

Did

242k23208443

edited Jul 23 at 11:02

Did

242k23208443

242k23208443

asked Jul 23 at 10:10

shayy

62

asked Jul 23 at 10:10

shayy

62

asked Jul 23 at 10:10

shayy

62

62

closed as off-topic by John Ma, Mathmo123, Taroccoesbrocco, Arnaud D., Asaf Karagila Aug 1 at 12:48

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Mathmo123, Taroccoesbrocco

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by John Ma, Mathmo123, Taroccoesbrocco, Arnaud D., Asaf Karagila Aug 1 at 12:48

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Mathmo123, Taroccoesbrocco

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, just write it out. What is $0@1$, for example? $0@7$? $0@b$?
  – lulu
  Jul 23 at 10:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  To prove $0$ is the identity, show that $a@0=0@a=a$ for an arbitrary choice of $a$
  – aidangallagher4
  Jul 23 at 10:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  To understand (and prove) in one strike all these properties (and how your TA built their exercise), note that $@$ is really the ordinary product in disguise, since $$(a@b)+1=(a+1)cdot(b+1)$$
  – Did
  Jul 23 at 11:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Possible duplicate of Proving $G$ is a group under a specific operation
  – Arnaud D.
  Aug 1 at 12:47
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, just write it out. What is $0@1$, for example? $0@7$? $0@b$?
  – lulu
  Jul 23 at 10:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  To prove $0$ is the identity, show that $a@0=0@a=a$ for an arbitrary choice of $a$
  – aidangallagher4
  Jul 23 at 10:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  To understand (and prove) in one strike all these properties (and how your TA built their exercise), note that $@$ is really the ordinary product in disguise, since $$(a@b)+1=(a+1)cdot(b+1)$$
  – Did
  Jul 23 at 11:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Possible duplicate of Proving $G$ is a group under a specific operation
  – Arnaud D.
  Aug 1 at 12:47
  

  
  
  
  

Well, just write it out. What is $0@1$, for example? $0@7$? $0@b$?
– lulu
Jul 23 at 10:15

Well, just write it out. What is $0@1$, for example? $0@7$? $0@b$?
– lulu
Jul 23 at 10:15

To prove $0$ is the identity, show that $a@0=0@a=a$ for an arbitrary choice of $a$
– aidangallagher4
Jul 23 at 10:16

To prove $0$ is the identity, show that $a@0=0@a=a$ for an arbitrary choice of $a$
– aidangallagher4
Jul 23 at 10:16

1

1

To understand (and prove) in one strike all these properties (and how your TA built their exercise), note that $@$ is really the ordinary product in disguise, since $$(a@b)+1=(a+1)cdot(b+1)$$
– Did
Jul 23 at 11:01

To understand (and prove) in one strike all these properties (and how your TA built their exercise), note that $@$ is really the ordinary product in disguise, since $$(a@b)+1=(a+1)cdot(b+1)$$
– Did
Jul 23 at 11:01

1

1

Possible duplicate of Proving $G$ is a group under a specific operation
– Arnaud D.
Aug 1 at 12:47

Possible duplicate of Proving $G$ is a group under a specific operation
– Arnaud D.
Aug 1 at 12:47

add a comment | 

2 Answers
2

active

oldest

votes

up vote
3
down vote

a) Show that 0 is an identity for the operation.

(I write $circ$ instead of @)

We have $acirc b:=ab+b+a$. Let $ainmathbbR$ be arbitrary, then:

$0circ a=0a+a+0=a$

$acirc 0=a0+0+a=a$

b) Show that some real numbers have inverses under the operation.

We have to find a pair $(a,b)inmathbbR^2$ sucht that $acirc b=0$ and $bcirc a=0$
First of all $acirc b=ab+a+b=ba+b+a=bcirc a$.

It has to be $ab+a+b=0$. $a(b+1)=-bLeftrightarrow a=frac-bb+1$ for $bneq -1$

Therefor $a$ has to be of the form $frac-bb+1$, to be an inverse of $bneq -1$.

$frac-bb+1b+b+frac-bb+1=frac-b^2-bb+1+fracb^2+bb+1=fracb^2-b^2+b-bb+1=0$

c) Find a counter-example to show that, for this operation inverses do not exist for all the real numbers.

For $a=-1$ there can not be an inverse.

$-1circ b=-b+b-1=-1$

Note, that it does not depend on $b$. So there is no chance, that it ever will be $0$.

share|cite|improve this answer

edited Jul 23 at 10:42

Bernard

110k635103

answered Jul 23 at 10:28

Cornman

2,41221127

add a comment | 

up vote
2
down vote

Note this operation is commutative, so $A$ has an inverse under this operation if and only if there exists $x$ such that $;a@x=ax+x+a=0$.

Can you determine for which $a$ this equation has a solution?

share|cite|improve this answer

answered Jul 23 at 10:19

Bernard

110k635103

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

a) Show that 0 is an identity for the operation.

(I write $circ$ instead of @)

We have $acirc b:=ab+b+a$. Let $ainmathbbR$ be arbitrary, then:

$0circ a=0a+a+0=a$

$acirc 0=a0+0+a=a$

b) Show that some real numbers have inverses under the operation.

We have to find a pair $(a,b)inmathbbR^2$ sucht that $acirc b=0$ and $bcirc a=0$
First of all $acirc b=ab+a+b=ba+b+a=bcirc a$.

It has to be $ab+a+b=0$. $a(b+1)=-bLeftrightarrow a=frac-bb+1$ for $bneq -1$

Therefor $a$ has to be of the form $frac-bb+1$, to be an inverse of $bneq -1$.

$frac-bb+1b+b+frac-bb+1=frac-b^2-bb+1+fracb^2+bb+1=fracb^2-b^2+b-bb+1=0$

c) Find a counter-example to show that, for this operation inverses do not exist for all the real numbers.

For $a=-1$ there can not be an inverse.

$-1circ b=-b+b-1=-1$

Note, that it does not depend on $b$. So there is no chance, that it ever will be $0$.

share|cite|improve this answer

edited Jul 23 at 10:42

Bernard

110k635103

answered Jul 23 at 10:28

Cornman

2,41221127

add a comment | 

up vote
3
down vote

a) Show that 0 is an identity for the operation.

(I write $circ$ instead of @)

We have $acirc b:=ab+b+a$. Let $ainmathbbR$ be arbitrary, then:

$0circ a=0a+a+0=a$

$acirc 0=a0+0+a=a$

b) Show that some real numbers have inverses under the operation.

We have to find a pair $(a,b)inmathbbR^2$ sucht that $acirc b=0$ and $bcirc a=0$
First of all $acirc b=ab+a+b=ba+b+a=bcirc a$.

It has to be $ab+a+b=0$. $a(b+1)=-bLeftrightarrow a=frac-bb+1$ for $bneq -1$

Therefor $a$ has to be of the form $frac-bb+1$, to be an inverse of $bneq -1$.

$frac-bb+1b+b+frac-bb+1=frac-b^2-bb+1+fracb^2+bb+1=fracb^2-b^2+b-bb+1=0$

c) Find a counter-example to show that, for this operation inverses do not exist for all the real numbers.

For $a=-1$ there can not be an inverse.

$-1circ b=-b+b-1=-1$

Note, that it does not depend on $b$. So there is no chance, that it ever will be $0$.

share|cite|improve this answer

edited Jul 23 at 10:42

Bernard

110k635103

answered Jul 23 at 10:28

Cornman

2,41221127

add a comment | 

up vote
3
down vote

up vote
3
down vote

a) Show that 0 is an identity for the operation.

(I write $circ$ instead of @)

We have $acirc b:=ab+b+a$. Let $ainmathbbR$ be arbitrary, then:

$0circ a=0a+a+0=a$

$acirc 0=a0+0+a=a$

b) Show that some real numbers have inverses under the operation.

We have to find a pair $(a,b)inmathbbR^2$ sucht that $acirc b=0$ and $bcirc a=0$
First of all $acirc b=ab+a+b=ba+b+a=bcirc a$.

It has to be $ab+a+b=0$. $a(b+1)=-bLeftrightarrow a=frac-bb+1$ for $bneq -1$

Therefor $a$ has to be of the form $frac-bb+1$, to be an inverse of $bneq -1$.

$frac-bb+1b+b+frac-bb+1=frac-b^2-bb+1+fracb^2+bb+1=fracb^2-b^2+b-bb+1=0$

c) Find a counter-example to show that, for this operation inverses do not exist for all the real numbers.

For $a=-1$ there can not be an inverse.

$-1circ b=-b+b-1=-1$

Note, that it does not depend on $b$. So there is no chance, that it ever will be $0$.

share|cite|improve this answer

edited Jul 23 at 10:42

Bernard

110k635103

answered Jul 23 at 10:28

Cornman

2,41221127

a) Show that 0 is an identity for the operation.

(I write $circ$ instead of @)

We have $acirc b:=ab+b+a$. Let $ainmathbbR$ be arbitrary, then:

$0circ a=0a+a+0=a$

$acirc 0=a0+0+a=a$

b) Show that some real numbers have inverses under the operation.

We have to find a pair $(a,b)inmathbbR^2$ sucht that $acirc b=0$ and $bcirc a=0$
First of all $acirc b=ab+a+b=ba+b+a=bcirc a$.

It has to be $ab+a+b=0$. $a(b+1)=-bLeftrightarrow a=frac-bb+1$ for $bneq -1$

Therefor $a$ has to be of the form $frac-bb+1$, to be an inverse of $bneq -1$.

$frac-bb+1b+b+frac-bb+1=frac-b^2-bb+1+fracb^2+bb+1=fracb^2-b^2+b-bb+1=0$

c) Find a counter-example to show that, for this operation inverses do not exist for all the real numbers.

For $a=-1$ there can not be an inverse.

$-1circ b=-b+b-1=-1$

Note, that it does not depend on $b$. So there is no chance, that it ever will be $0$.

share|cite|improve this answer

edited Jul 23 at 10:42

Bernard

110k635103

answered Jul 23 at 10:28

Cornman

2,41221127

share|cite|improve this answer

share|cite|improve this answer

edited Jul 23 at 10:42

Bernard

110k635103

edited Jul 23 at 10:42

Bernard

110k635103

edited Jul 23 at 10:42

Bernard

110k635103

110k635103

answered Jul 23 at 10:28

Cornman

2,41221127

answered Jul 23 at 10:28

Cornman

2,41221127

answered Jul 23 at 10:28

Cornman

2,41221127

2,41221127

add a comment | 

add a comment | 

up vote
2
down vote

Note this operation is commutative, so $A$ has an inverse under this operation if and only if there exists $x$ such that $;a@x=ax+x+a=0$.

Can you determine for which $a$ this equation has a solution?

share|cite|improve this answer

answered Jul 23 at 10:19

Bernard

110k635103

add a comment | 

up vote
2
down vote

Note this operation is commutative, so $A$ has an inverse under this operation if and only if there exists $x$ such that $;a@x=ax+x+a=0$.

Can you determine for which $a$ this equation has a solution?

share|cite|improve this answer

answered Jul 23 at 10:19

Bernard

110k635103

add a comment | 

up vote
2
down vote

up vote
2
down vote

Note this operation is commutative, so $A$ has an inverse under this operation if and only if there exists $x$ such that $;a@x=ax+x+a=0$.

Can you determine for which $a$ this equation has a solution?

share|cite|improve this answer

answered Jul 23 at 10:19

Bernard

110k635103

Note this operation is commutative, so $A$ has an inverse under this operation if and only if there exists $x$ such that $;a@x=ax+x+a=0$.

Can you determine for which $a$ this equation has a solution?

share|cite|improve this answer

answered Jul 23 at 10:19

Bernard

110k635103

share|cite|improve this answer

share|cite|improve this answer

answered Jul 23 at 10:19

Bernard

110k635103

answered Jul 23 at 10:19

Bernard

110k635103

answered Jul 23 at 10:19

Bernard

110k635103

110k635103

add a comment | 

add a comment |