What are the main points to remember to solve these kinds of problems? [closed]

Clash Royale CLAN TAG#URR8PPP

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Problem #$1$

If $ab>0$, then which one of the following must be true?

$(A)$ $a/b>0 $

$(B)$ $a-b>0 $

$(C)$ $a+b>0 $

$(D)$ $b-a>0 $

$(E)$ $a+b<0 $

Problem #$2$

If $x+z > y+z$ then which of the following must be true?

$(A)$ $x-z>y-z$

$(B)$ $xz>yz$

$(C)$ $x/z>y/z$

$(D)$ $x/2>y/z$

$(E)$ $2x+z>2y+z$

What are the main points to remember to solve these kinds of problems?

gre-exam

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edited Jul 15 at 5:17

asked Jul 15 at 4:18

yahoo.com

391216

closed as off-topic by Alex Francisco, anomaly, John Ma, Mostafa Ayaz, user223391 Jul 15 at 21:29

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." – Alex Francisco, anomaly, John Ma, Mostafa Ayaz, Community

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are these real numbers?
  – zhw.
  Jul 15 at 7:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @zhw., yes. ....
  – yahoo.com
  Jul 15 at 8:11
  

  
  
  
  

add a comment | 

up vote
-2
down vote

favorite

Problem #$1$

If $ab>0$, then which one of the following must be true?

$(A)$ $a/b>0 $

$(B)$ $a-b>0 $

$(C)$ $a+b>0 $

$(D)$ $b-a>0 $

$(E)$ $a+b<0 $

Problem #$2$

If $x+z > y+z$ then which of the following must be true?

$(A)$ $x-z>y-z$

$(B)$ $xz>yz$

$(C)$ $x/z>y/z$

$(D)$ $x/2>y/z$

$(E)$ $2x+z>2y+z$

What are the main points to remember to solve these kinds of problems?

gre-exam

share|cite|improve this question

edited Jul 15 at 5:17

asked Jul 15 at 4:18

yahoo.com

391216

closed as off-topic by Alex Francisco, anomaly, John Ma, Mostafa Ayaz, user223391 Jul 15 at 21:29

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." – Alex Francisco, anomaly, John Ma, Mostafa Ayaz, Community

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are these real numbers?
  – zhw.
  Jul 15 at 7:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @zhw., yes. ....
  – yahoo.com
  Jul 15 at 8:11
  

  
  
  
  

add a comment | 

up vote
-2
down vote

favorite

up vote
-2
down vote

favorite

Problem #$1$

If $ab>0$, then which one of the following must be true?

$(A)$ $a/b>0 $

$(B)$ $a-b>0 $

$(C)$ $a+b>0 $

$(D)$ $b-a>0 $

$(E)$ $a+b<0 $

Problem #$2$

If $x+z > y+z$ then which of the following must be true?

$(A)$ $x-z>y-z$

$(B)$ $xz>yz$

$(C)$ $x/z>y/z$

$(D)$ $x/2>y/z$

$(E)$ $2x+z>2y+z$

What are the main points to remember to solve these kinds of problems?

gre-exam

share|cite|improve this question

edited Jul 15 at 5:17

asked Jul 15 at 4:18

yahoo.com

391216

Problem #$1$

If $ab>0$, then which one of the following must be true?

$(A)$ $a/b>0 $

$(B)$ $a-b>0 $

$(C)$ $a+b>0 $

$(D)$ $b-a>0 $

$(E)$ $a+b<0 $

Problem #$2$

If $x+z > y+z$ then which of the following must be true?

$(A)$ $x-z>y-z$

$(B)$ $xz>yz$

$(C)$ $x/z>y/z$

$(D)$ $x/2>y/z$

$(E)$ $2x+z>2y+z$

What are the main points to remember to solve these kinds of problems?

gre-exam

share|cite|improve this question

edited Jul 15 at 5:17

asked Jul 15 at 4:18

yahoo.com

391216

share|cite|improve this question

share|cite|improve this question

edited Jul 15 at 5:17

edited Jul 15 at 5:17

edited Jul 15 at 5:17

asked Jul 15 at 4:18

yahoo.com

391216

asked Jul 15 at 4:18

yahoo.com

391216

asked Jul 15 at 4:18

yahoo.com

391216

391216

closed as off-topic by Alex Francisco, anomaly, John Ma, Mostafa Ayaz, user223391 Jul 15 at 21:29

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." – Alex Francisco, anomaly, John Ma, Mostafa Ayaz, Community

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

closed as off-topic by Alex Francisco, anomaly, John Ma, Mostafa Ayaz, user223391 Jul 15 at 21:29

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." – Alex Francisco, anomaly, John Ma, Mostafa Ayaz, Community

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are these real numbers?
  – zhw.
  Jul 15 at 7:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @zhw., yes. ....
  – yahoo.com
  Jul 15 at 8:11
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are these real numbers?
  – zhw.
  Jul 15 at 7:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @zhw., yes. ....
  – yahoo.com
  Jul 15 at 8:11
  

  
  
  
  

Are these real numbers?
– zhw.
Jul 15 at 7:15

Are these real numbers?
– zhw.
Jul 15 at 7:15

@zhw., yes. ....
– yahoo.com
Jul 15 at 8:11

@zhw., yes. ....
– yahoo.com
Jul 15 at 8:11

add a comment | 

2 Answers
2

active

oldest

votes

up vote
1
down vote



accepted

You are dealing with an ordered field $F$ if there is a set $P$ (the P corresponds with "positive") such that the $3$ sets: $$P text, 0text and -P$$ form a partition (i.e. the sets are not-empty, are disjoint and cover $F$).

Here $-P:=-xmid xin P$

This with:$$forall x,yin P[x+yin Ptext and xyin P]$$i.e. the positive set $P$ is closed under addition and multiplication.

An order $<$ on $F$ is induced by:$$x<ytext if y-xin P$$

Equipped with this knowledge problems like #1 and #2 can always be solved by you.

---

Edit:

It is handsome to start with the observation that $1in P$, which is no extra knowledge, but is something that can be proved on the knowledge mentioned above (so is no extra knowledge).

Proof:

Suppose that $1notin P$.

Then - because also $1notin0$ we must have $1in-P$ or equivalently $-1in P$.

But $P$ is closed under multiplication allowing us to conclude that $1=(-1)(-1)in P$ and a contradiction is found, so we conclude that $1in P$.

share|cite|improve this answer

edited Jul 15 at 5:00

answered Jul 15 at 4:42

drhab

86.7k541118

add a comment | 

up vote
1
down vote

What are the main points to remember to solve these kinds of problems?

I'll answer this by sharing my thought process when solving these.

#1

If $ab > 0$, then the product of those two numbers is positive. That means either $a$ and $b$ are both positive or both negative. If I have two numbers with the same sign (i.e. they are both positive or both negative), can I say anything definitively about their sum, difference, or quotient?

Final hint: If I have two numbers with the same sign (i.e. they are both positive or both negative), what can I say about their quotient?

#2

If $x+z>y+z$, then $x>y$. Can this inequality be manipulated to match one of the answer choices?

Here's a narrative hint: suppose we are both in a race, and I am ahead of you. If we both walk backwards the same distance, what is my distance ran compared to yours?

share|cite|improve this answer

answered Jul 15 at 4:41

Tiwa Aina

2,576319

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

You are dealing with an ordered field $F$ if there is a set $P$ (the P corresponds with "positive") such that the $3$ sets: $$P text, 0text and -P$$ form a partition (i.e. the sets are not-empty, are disjoint and cover $F$).

Here $-P:=-xmid xin P$

This with:$$forall x,yin P[x+yin Ptext and xyin P]$$i.e. the positive set $P$ is closed under addition and multiplication.

An order $<$ on $F$ is induced by:$$x<ytext if y-xin P$$

Equipped with this knowledge problems like #1 and #2 can always be solved by you.

---

Edit:

It is handsome to start with the observation that $1in P$, which is no extra knowledge, but is something that can be proved on the knowledge mentioned above (so is no extra knowledge).

Proof:

Suppose that $1notin P$.

Then - because also $1notin0$ we must have $1in-P$ or equivalently $-1in P$.

But $P$ is closed under multiplication allowing us to conclude that $1=(-1)(-1)in P$ and a contradiction is found, so we conclude that $1in P$.

share|cite|improve this answer

edited Jul 15 at 5:00

answered Jul 15 at 4:42

drhab

86.7k541118

add a comment | 

up vote
1
down vote



accepted

You are dealing with an ordered field $F$ if there is a set $P$ (the P corresponds with "positive") such that the $3$ sets: $$P text, 0text and -P$$ form a partition (i.e. the sets are not-empty, are disjoint and cover $F$).

Here $-P:=-xmid xin P$

This with:$$forall x,yin P[x+yin Ptext and xyin P]$$i.e. the positive set $P$ is closed under addition and multiplication.

An order $<$ on $F$ is induced by:$$x<ytext if y-xin P$$

Equipped with this knowledge problems like #1 and #2 can always be solved by you.

---

Edit:

It is handsome to start with the observation that $1in P$, which is no extra knowledge, but is something that can be proved on the knowledge mentioned above (so is no extra knowledge).

Proof:

Suppose that $1notin P$.

Then - because also $1notin0$ we must have $1in-P$ or equivalently $-1in P$.

But $P$ is closed under multiplication allowing us to conclude that $1=(-1)(-1)in P$ and a contradiction is found, so we conclude that $1in P$.

share|cite|improve this answer

edited Jul 15 at 5:00

answered Jul 15 at 4:42

drhab

86.7k541118

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

You are dealing with an ordered field $F$ if there is a set $P$ (the P corresponds with "positive") such that the $3$ sets: $$P text, 0text and -P$$ form a partition (i.e. the sets are not-empty, are disjoint and cover $F$).

Here $-P:=-xmid xin P$

This with:$$forall x,yin P[x+yin Ptext and xyin P]$$i.e. the positive set $P$ is closed under addition and multiplication.

An order $<$ on $F$ is induced by:$$x<ytext if y-xin P$$

Equipped with this knowledge problems like #1 and #2 can always be solved by you.

---

Edit:

It is handsome to start with the observation that $1in P$, which is no extra knowledge, but is something that can be proved on the knowledge mentioned above (so is no extra knowledge).

Proof:

Suppose that $1notin P$.

Then - because also $1notin0$ we must have $1in-P$ or equivalently $-1in P$.

But $P$ is closed under multiplication allowing us to conclude that $1=(-1)(-1)in P$ and a contradiction is found, so we conclude that $1in P$.

share|cite|improve this answer

edited Jul 15 at 5:00

answered Jul 15 at 4:42

drhab

86.7k541118

You are dealing with an ordered field $F$ if there is a set $P$ (the P corresponds with "positive") such that the $3$ sets: $$P text, 0text and -P$$ form a partition (i.e. the sets are not-empty, are disjoint and cover $F$).

Here $-P:=-xmid xin P$

This with:$$forall x,yin P[x+yin Ptext and xyin P]$$i.e. the positive set $P$ is closed under addition and multiplication.

An order $<$ on $F$ is induced by:$$x<ytext if y-xin P$$

Equipped with this knowledge problems like #1 and #2 can always be solved by you.

---

Edit:

It is handsome to start with the observation that $1in P$, which is no extra knowledge, but is something that can be proved on the knowledge mentioned above (so is no extra knowledge).

Proof:

Suppose that $1notin P$.

Then - because also $1notin0$ we must have $1in-P$ or equivalently $-1in P$.

But $P$ is closed under multiplication allowing us to conclude that $1=(-1)(-1)in P$ and a contradiction is found, so we conclude that $1in P$.

share|cite|improve this answer

edited Jul 15 at 5:00

answered Jul 15 at 4:42

drhab

86.7k541118

share|cite|improve this answer

share|cite|improve this answer

edited Jul 15 at 5:00

edited Jul 15 at 5:00

edited Jul 15 at 5:00

answered Jul 15 at 4:42

drhab

86.7k541118

answered Jul 15 at 4:42

drhab

86.7k541118

answered Jul 15 at 4:42

drhab

86.7k541118

86.7k541118

add a comment | 

add a comment | 

up vote
1
down vote

What are the main points to remember to solve these kinds of problems?

I'll answer this by sharing my thought process when solving these.

#1

If $ab > 0$, then the product of those two numbers is positive. That means either $a$ and $b$ are both positive or both negative. If I have two numbers with the same sign (i.e. they are both positive or both negative), can I say anything definitively about their sum, difference, or quotient?

Final hint: If I have two numbers with the same sign (i.e. they are both positive or both negative), what can I say about their quotient?

#2

If $x+z>y+z$, then $x>y$. Can this inequality be manipulated to match one of the answer choices?

Here's a narrative hint: suppose we are both in a race, and I am ahead of you. If we both walk backwards the same distance, what is my distance ran compared to yours?

share|cite|improve this answer

answered Jul 15 at 4:41

Tiwa Aina

2,576319

add a comment | 

up vote
1
down vote

What are the main points to remember to solve these kinds of problems?

I'll answer this by sharing my thought process when solving these.

#1

If $ab > 0$, then the product of those two numbers is positive. That means either $a$ and $b$ are both positive or both negative. If I have two numbers with the same sign (i.e. they are both positive or both negative), can I say anything definitively about their sum, difference, or quotient?

Final hint: If I have two numbers with the same sign (i.e. they are both positive or both negative), what can I say about their quotient?

#2

If $x+z>y+z$, then $x>y$. Can this inequality be manipulated to match one of the answer choices?

Here's a narrative hint: suppose we are both in a race, and I am ahead of you. If we both walk backwards the same distance, what is my distance ran compared to yours?

share|cite|improve this answer

answered Jul 15 at 4:41

Tiwa Aina

2,576319

add a comment | 

up vote
1
down vote

up vote
1
down vote

What are the main points to remember to solve these kinds of problems?

I'll answer this by sharing my thought process when solving these.

#1

If $ab > 0$, then the product of those two numbers is positive. That means either $a$ and $b$ are both positive or both negative. If I have two numbers with the same sign (i.e. they are both positive or both negative), can I say anything definitively about their sum, difference, or quotient?

Final hint: If I have two numbers with the same sign (i.e. they are both positive or both negative), what can I say about their quotient?

#2

If $x+z>y+z$, then $x>y$. Can this inequality be manipulated to match one of the answer choices?

Here's a narrative hint: suppose we are both in a race, and I am ahead of you. If we both walk backwards the same distance, what is my distance ran compared to yours?

share|cite|improve this answer

answered Jul 15 at 4:41

Tiwa Aina

2,576319

What are the main points to remember to solve these kinds of problems?

I'll answer this by sharing my thought process when solving these.

#1

If $ab > 0$, then the product of those two numbers is positive. That means either $a$ and $b$ are both positive or both negative. If I have two numbers with the same sign (i.e. they are both positive or both negative), can I say anything definitively about their sum, difference, or quotient?

Final hint: If I have two numbers with the same sign (i.e. they are both positive or both negative), what can I say about their quotient?

#2

If $x+z>y+z$, then $x>y$. Can this inequality be manipulated to match one of the answer choices?

Here's a narrative hint: suppose we are both in a race, and I am ahead of you. If we both walk backwards the same distance, what is my distance ran compared to yours?

share|cite|improve this answer

answered Jul 15 at 4:41

Tiwa Aina

2,576319

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 4:41

Tiwa Aina

2,576319

answered Jul 15 at 4:41

Tiwa Aina

2,576319

answered Jul 15 at 4:41

Tiwa Aina

2,576319

2,576319

add a comment | 

add a comment |