Mixing
Prove that if $a$ and $b$ are some integers, then $a^2 - 4b$ does not equal $2$.
$begingroup$
Prove that if $a$ and $b$ are some integers, then $a^2 - 4b$ does not equal $2$.
How do I prove this, I tried using contradiction but I'm not sure how I can contradict this statement?
integers
edited Oct 23 '16 at 1:22
Em.
15k72037
asked Oct 23 '16 at 0:07
Chance GordonChance Gordon
1069
$endgroup$
add a comment |
$begingroup$
Prove that if $a$ and $b$ are some integers, then $a^2 - 4b$ does not equal $2$.
How do I prove this, I tried using contradiction but I'm not sure how I can contradict this statement?
integers
edited Oct 23 '16 at 1:22
Em.
15k72037
asked Oct 23 '16 at 0:07
Chance GordonChance Gordon
1069
$endgroup$
$begingroup$
Is it possile if a is odd? Is it possible if a is even.
$endgroup$
– fleablood
Oct 23 '16 at 0:11
$begingroup$
It doesnt say anything about it being odd or even in the question
$endgroup$
– Chance Gordon
Oct 23 '16 at 0:16
$begingroup$
So? That doesn't mean you can't say anything about it being odd or even. So, again. What happens if a is odd? What happens if a is even?
$endgroup$
– fleablood
Oct 23 '16 at 0:37
$begingroup$
Contradiction: Let $a^2 - 4b = 2$ then $a^2 = 4b + 2 = 2(2b + 1)$ so $a$ is even. Let $a = 2c$. $4c^2 = 2(2b + 1)$ so $2c^2 = 2b + 1$. Hmmmmm.....
$endgroup$
– fleablood
Oct 23 '16 at 0:40
add a comment |
$begingroup$
Prove that if $a$ and $b$ are some integers, then $a^2 - 4b$ does not equal $2$.
How do I prove this, I tried using contradiction but I'm not sure how I can contradict this statement?
integers
edited Oct 23 '16 at 1:22
Em.
15k72037
asked Oct 23 '16 at 0:07
Chance GordonChance Gordon
1069
$endgroup$
Prove that if $a$ and $b$ are some integers, then $a^2 - 4b$ does not equal $2$.
How do I prove this, I tried using contradiction but I'm not sure how I can contradict this statement?
integers
integers
edited Oct 23 '16 at 1:22
Em.
15k72037
asked Oct 23 '16 at 0:07
Chance GordonChance Gordon
1069
edited Oct 23 '16 at 1:22
Em.
15k72037
asked Oct 23 '16 at 0:07
Chance GordonChance Gordon
1069
edited Oct 23 '16 at 1:22
Em.
15k72037
edited Oct 23 '16 at 1:22
Em.
15k72037
edited Oct 23 '16 at 1:22
Em.
15k72037
15k72037
asked Oct 23 '16 at 0:07
Chance GordonChance Gordon
1069
asked Oct 23 '16 at 0:07
Chance GordonChance Gordon
1069
asked Oct 23 '16 at 0:07
Chance GordonChance Gordon
1069
1069
$begingroup$
Is it possile if a is odd? Is it possible if a is even.
$endgroup$
– fleablood
Oct 23 '16 at 0:11
$begingroup$
It doesnt say anything about it being odd or even in the question
$endgroup$
– Chance Gordon
Oct 23 '16 at 0:16
$begingroup$
So? That doesn't mean you can't say anything about it being odd or even. So, again. What happens if a is odd? What happens if a is even?
$endgroup$
– fleablood
Oct 23 '16 at 0:37
$begingroup$
Contradiction: Let $a^2 - 4b = 2$ then $a^2 = 4b + 2 = 2(2b + 1)$ so $a$ is even. Let $a = 2c$. $4c^2 = 2(2b + 1)$ so $2c^2 = 2b + 1$. Hmmmmm.....
$endgroup$
– fleablood
Oct 23 '16 at 0:40
add a comment |
$begingroup$
Is it possile if a is odd? Is it possible if a is even.
$endgroup$
– fleablood
Oct 23 '16 at 0:11
$begingroup$
It doesnt say anything about it being odd or even in the question
$endgroup$
– Chance Gordon
Oct 23 '16 at 0:16
$begingroup$
So? That doesn't mean you can't say anything about it being odd or even. So, again. What happens if a is odd? What happens if a is even?
$endgroup$
– fleablood
Oct 23 '16 at 0:37
$begingroup$
Contradiction: Let $a^2 - 4b = 2$ then $a^2 = 4b + 2 = 2(2b + 1)$ so $a$ is even. Let $a = 2c$. $4c^2 = 2(2b + 1)$ so $2c^2 = 2b + 1$. Hmmmmm.....
$endgroup$
– fleablood
Oct 23 '16 at 0:40
$begingroup$
Is it possile if a is odd? Is it possible if a is even.
$endgroup$
– fleablood
Oct 23 '16 at 0:11
$begingroup$
Is it possile if a is odd? Is it possible if a is even.
$endgroup$
– fleablood
Oct 23 '16 at 0:11
$begingroup$
It doesnt say anything about it being odd or even in the question
$endgroup$
– Chance Gordon
Oct 23 '16 at 0:16
$begingroup$
It doesnt say anything about it being odd or even in the question
$endgroup$
– Chance Gordon
Oct 23 '16 at 0:16
$begingroup$
So? That doesn't mean you can't say anything about it being odd or even. So, again. What happens if a is odd? What happens if a is even?
$endgroup$
– fleablood
Oct 23 '16 at 0:37
$begingroup$
So? That doesn't mean you can't say anything about it being odd or even. So, again. What happens if a is odd? What happens if a is even?
$endgroup$
– fleablood
Oct 23 '16 at 0:37
$begingroup$
Contradiction: Let $a^2 - 4b = 2$ then $a^2 = 4b + 2 = 2(2b + 1)$ so $a$ is even. Let $a = 2c$. $4c^2 = 2(2b + 1)$ so $2c^2 = 2b + 1$. Hmmmmm.....
$endgroup$
– fleablood
Oct 23 '16 at 0:40
$begingroup$
Contradiction: Let $a^2 - 4b = 2$ then $a^2 = 4b + 2 = 2(2b + 1)$ so $a$ is even. Let $a = 2c$. $4c^2 = 2(2b + 1)$ so $2c^2 = 2b + 1$. Hmmmmm.....
$endgroup$
– fleablood
Oct 23 '16 at 0:40
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint:
Use congruences. Such an equality would imply $2$ is a square modulo $4$. Write the list of all squares modulo $4$ and see if $2$ is among them.
Some details:
Remember two integers are congruent modulo $4$ if they have the same remainder when divided by $4$. This is equivalent to ‘one of them is equal to the other, pus a multiple of $4$’.
So $a^2-4b=2iff a^2=2+4biff a^2equiv 2mod 4$.
Now the classes of congruence modulo $4$ are $0$, $pm1$, $2$, and the relation of congruence is compatible with addition and multiplication. Thus the clsses of congruence of the squares are
- $0^2=0$;
- $(pm1)^2=1$;
- $2^2=4equiv0$.
Neither $-1$ nor $2$ are in the list of (congruence classes) of squares modulo 4: we proved $2$ and $-1$ are not squares modulo $4$.
answered Oct 23 '16 at 0:19
BernardBernard
122k741116
$endgroup$
$begingroup$
I have little understanding of modulos and where and how to use them. Could you explain a bit?
$endgroup$
– Chance Gordon
Oct 23 '16 at 0:23
$begingroup$
@Chance Gordon: I've added details. Is that clearer?
$endgroup$
– Bernard
Oct 23 '16 at 0:35
add a comment |
$begingroup$
Your inequality can be rewritten as $a^2neq 4b+2= 2(2b+1)$.
Note that $2b+1$ is always an odd number whatever be $b$.
SO your question boils down to showing $a^2$ can never be double an odd number.
When $a$ is odd its square cannot be double of any number, being odd again.
When $a$ is even $ a^2$ will be an even number squared, that is $2mtimes 2m$ which is double of an even number. Done.
answered Oct 23 '16 at 1:27
P VanchinathanP Vanchinathan
15.3k12136
$endgroup$
add a comment |
$begingroup$
Contradiction: Write as $(a+2b)(a-2b)=2$. Since $a$ and $b$ are integers, $a+2b$ and $a-2b$ are also integers. Therefore either $a+2b=2$ and $a-2b=1$, or $a+2b=1$ and $a-2b=2$ (or $-2$ and $-1$ or $-1$ and $-2$). There are $4$ pairs of simultaneous equations that could be true, however solving gives either: $a=3/2$, $b=pm 1/4$ or $a=-3/2$, $b=pm 1/4$. Any of these solutions contradicts that $a$ and $b$ were integers.
edited Jan 21 at 23:52
idriskameni
757321
answered Jan 21 at 22:25
Jonty FreemanJonty Freeman
1
$endgroup$
$begingroup$
$(a+2b)(a-2b) = a^2 - 4b^2 ne a^2 - 4b$.
$endgroup$
– fleablood
Jan 21 at 22:53
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Use congruences. Such an equality would imply $2$ is a square modulo $4$. Write the list of all squares modulo $4$ and see if $2$ is among them.
Some details:
Remember two integers are congruent modulo $4$ if they have the same remainder when divided by $4$. This is equivalent to ‘one of them is equal to the other, pus a multiple of $4$’.
So $a^2-4b=2iff a^2=2+4biff a^2equiv 2mod 4$.
Now the classes of congruence modulo $4$ are $0$, $pm1$, $2$, and the relation of congruence is compatible with addition and multiplication. Thus the clsses of congruence of the squares are
- $0^2=0$;
- $(pm1)^2=1$;
- $2^2=4equiv0$.
Neither $-1$ nor $2$ are in the list of (congruence classes) of squares modulo 4: we proved $2$ and $-1$ are not squares modulo $4$.
answered Oct 23 '16 at 0:19
BernardBernard
122k741116
$endgroup$
$begingroup$
I have little understanding of modulos and where and how to use them. Could you explain a bit?
$endgroup$
– Chance Gordon
Oct 23 '16 at 0:23
$begingroup$
@Chance Gordon: I've added details. Is that clearer?
$endgroup$
– Bernard
Oct 23 '16 at 0:35
add a comment |
$begingroup$
Hint:
Use congruences. Such an equality would imply $2$ is a square modulo $4$. Write the list of all squares modulo $4$ and see if $2$ is among them.
Some details:
Remember two integers are congruent modulo $4$ if they have the same remainder when divided by $4$. This is equivalent to ‘one of them is equal to the other, pus a multiple of $4$’.
So $a^2-4b=2iff a^2=2+4biff a^2equiv 2mod 4$.
Now the classes of congruence modulo $4$ are $0$, $pm1$, $2$, and the relation of congruence is compatible with addition and multiplication. Thus the clsses of congruence of the squares are
- $0^2=0$;
- $(pm1)^2=1$;
- $2^2=4equiv0$.
Neither $-1$ nor $2$ are in the list of (congruence classes) of squares modulo 4: we proved $2$ and $-1$ are not squares modulo $4$.
answered Oct 23 '16 at 0:19
BernardBernard
122k741116
$endgroup$
$begingroup$
I have little understanding of modulos and where and how to use them. Could you explain a bit?
$endgroup$
– Chance Gordon
Oct 23 '16 at 0:23
$begingroup$
@Chance Gordon: I've added details. Is that clearer?
$endgroup$
– Bernard
Oct 23 '16 at 0:35
add a comment |
$begingroup$
Hint:
Use congruences. Such an equality would imply $2$ is a square modulo $4$. Write the list of all squares modulo $4$ and see if $2$ is among them.
Some details:
Remember two integers are congruent modulo $4$ if they have the same remainder when divided by $4$. This is equivalent to ‘one of them is equal to the other, pus a multiple of $4$’.
So $a^2-4b=2iff a^2=2+4biff a^2equiv 2mod 4$.
Now the classes of congruence modulo $4$ are $0$, $pm1$, $2$, and the relation of congruence is compatible with addition and multiplication. Thus the clsses of congruence of the squares are
- $0^2=0$;
- $(pm1)^2=1$;
- $2^2=4equiv0$.
Neither $-1$ nor $2$ are in the list of (congruence classes) of squares modulo 4: we proved $2$ and $-1$ are not squares modulo $4$.
answered Oct 23 '16 at 0:19
BernardBernard
122k741116
$endgroup$
Hint:
Use congruences. Such an equality would imply $2$ is a square modulo $4$. Write the list of all squares modulo $4$ and see if $2$ is among them.
Some details:
Remember two integers are congruent modulo $4$ if they have the same remainder when divided by $4$. This is equivalent to ‘one of them is equal to the other, pus a multiple of $4$’.
So $a^2-4b=2iff a^2=2+4biff a^2equiv 2mod 4$.
Now the classes of congruence modulo $4$ are $0$, $pm1$, $2$, and the relation of congruence is compatible with addition and multiplication. Thus the clsses of congruence of the squares are
- $0^2=0$;
- $(pm1)^2=1$;
- $2^2=4equiv0$.
Neither $-1$ nor $2$ are in the list of (congruence classes) of squares modulo 4: we proved $2$ and $-1$ are not squares modulo $4$.
answered Oct 23 '16 at 0:19
BernardBernard
122k741116
edited Oct 23 '16 at 0:34
answered Oct 23 '16 at 0:19
BernardBernard
122k741116
answered Oct 23 '16 at 0:19
BernardBernard
122k741116
answered Oct 23 '16 at 0:19
BernardBernard
122k741116
122k741116
$begingroup$
I have little understanding of modulos and where and how to use them. Could you explain a bit?
$endgroup$
– Chance Gordon
Oct 23 '16 at 0:23
$begingroup$
@Chance Gordon: I've added details. Is that clearer?
$endgroup$
– Bernard
Oct 23 '16 at 0:35
add a comment |
$begingroup$
I have little understanding of modulos and where and how to use them. Could you explain a bit?
$endgroup$
– Chance Gordon
Oct 23 '16 at 0:23
$begingroup$
@Chance Gordon: I've added details. Is that clearer?
$endgroup$
– Bernard
Oct 23 '16 at 0:35
$begingroup$
I have little understanding of modulos and where and how to use them. Could you explain a bit?
$endgroup$
– Chance Gordon
Oct 23 '16 at 0:23
$begingroup$
I have little understanding of modulos and where and how to use them. Could you explain a bit?
$endgroup$
– Chance Gordon
Oct 23 '16 at 0:23
$begingroup$
@Chance Gordon: I've added details. Is that clearer?
$endgroup$
– Bernard
Oct 23 '16 at 0:35
$begingroup$
@Chance Gordon: I've added details. Is that clearer?
$endgroup$
– Bernard
Oct 23 '16 at 0:35
add a comment |
$begingroup$
Your inequality can be rewritten as $a^2neq 4b+2= 2(2b+1)$.
Note that $2b+1$ is always an odd number whatever be $b$.
SO your question boils down to showing $a^2$ can never be double an odd number.
When $a$ is odd its square cannot be double of any number, being odd again.
When $a$ is even $ a^2$ will be an even number squared, that is $2mtimes 2m$ which is double of an even number. Done.
answered Oct 23 '16 at 1:27
P VanchinathanP Vanchinathan
15.3k12136
$endgroup$
add a comment |
$begingroup$
Your inequality can be rewritten as $a^2neq 4b+2= 2(2b+1)$.
Note that $2b+1$ is always an odd number whatever be $b$.
SO your question boils down to showing $a^2$ can never be double an odd number.
When $a$ is odd its square cannot be double of any number, being odd again.
When $a$ is even $ a^2$ will be an even number squared, that is $2mtimes 2m$ which is double of an even number. Done.
answered Oct 23 '16 at 1:27
P VanchinathanP Vanchinathan
15.3k12136
$endgroup$
add a comment |
$begingroup$
Your inequality can be rewritten as $a^2neq 4b+2= 2(2b+1)$.
Note that $2b+1$ is always an odd number whatever be $b$.
SO your question boils down to showing $a^2$ can never be double an odd number.
When $a$ is odd its square cannot be double of any number, being odd again.
When $a$ is even $ a^2$ will be an even number squared, that is $2mtimes 2m$ which is double of an even number. Done.
answered Oct 23 '16 at 1:27
P VanchinathanP Vanchinathan
15.3k12136
$endgroup$
Your inequality can be rewritten as $a^2neq 4b+2= 2(2b+1)$.
Note that $2b+1$ is always an odd number whatever be $b$.
SO your question boils down to showing $a^2$ can never be double an odd number.
When $a$ is odd its square cannot be double of any number, being odd again.
When $a$ is even $ a^2$ will be an even number squared, that is $2mtimes 2m$ which is double of an even number. Done.
answered Oct 23 '16 at 1:27
P VanchinathanP Vanchinathan
15.3k12136
answered Oct 23 '16 at 1:27
P VanchinathanP Vanchinathan
15.3k12136
answered Oct 23 '16 at 1:27
P VanchinathanP Vanchinathan
15.3k12136
answered Oct 23 '16 at 1:27
P VanchinathanP Vanchinathan
15.3k12136
15.3k12136
add a comment |
add a comment |
$begingroup$
Contradiction: Write as $(a+2b)(a-2b)=2$. Since $a$ and $b$ are integers, $a+2b$ and $a-2b$ are also integers. Therefore either $a+2b=2$ and $a-2b=1$, or $a+2b=1$ and $a-2b=2$ (or $-2$ and $-1$ or $-1$ and $-2$). There are $4$ pairs of simultaneous equations that could be true, however solving gives either: $a=3/2$, $b=pm 1/4$ or $a=-3/2$, $b=pm 1/4$. Any of these solutions contradicts that $a$ and $b$ were integers.
edited Jan 21 at 23:52
idriskameni
757321
answered Jan 21 at 22:25
Jonty FreemanJonty Freeman
1
$endgroup$
$begingroup$
$(a+2b)(a-2b) = a^2 - 4b^2 ne a^2 - 4b$.
$endgroup$
– fleablood
Jan 21 at 22:53
add a comment |
$begingroup$
Contradiction: Write as $(a+2b)(a-2b)=2$. Since $a$ and $b$ are integers, $a+2b$ and $a-2b$ are also integers. Therefore either $a+2b=2$ and $a-2b=1$, or $a+2b=1$ and $a-2b=2$ (or $-2$ and $-1$ or $-1$ and $-2$). There are $4$ pairs of simultaneous equations that could be true, however solving gives either: $a=3/2$, $b=pm 1/4$ or $a=-3/2$, $b=pm 1/4$. Any of these solutions contradicts that $a$ and $b$ were integers.
edited Jan 21 at 23:52
idriskameni
757321
answered Jan 21 at 22:25
Jonty FreemanJonty Freeman
1
$endgroup$
$begingroup$
$(a+2b)(a-2b) = a^2 - 4b^2 ne a^2 - 4b$.
$endgroup$
– fleablood
Jan 21 at 22:53
add a comment |
$begingroup$
Contradiction: Write as $(a+2b)(a-2b)=2$. Since $a$ and $b$ are integers, $a+2b$ and $a-2b$ are also integers. Therefore either $a+2b=2$ and $a-2b=1$, or $a+2b=1$ and $a-2b=2$ (or $-2$ and $-1$ or $-1$ and $-2$). There are $4$ pairs of simultaneous equations that could be true, however solving gives either: $a=3/2$, $b=pm 1/4$ or $a=-3/2$, $b=pm 1/4$. Any of these solutions contradicts that $a$ and $b$ were integers.
edited Jan 21 at 23:52
idriskameni
757321
answered Jan 21 at 22:25
Jonty FreemanJonty Freeman
1
$endgroup$
Contradiction: Write as $(a+2b)(a-2b)=2$. Since $a$ and $b$ are integers, $a+2b$ and $a-2b$ are also integers. Therefore either $a+2b=2$ and $a-2b=1$, or $a+2b=1$ and $a-2b=2$ (or $-2$ and $-1$ or $-1$ and $-2$). There are $4$ pairs of simultaneous equations that could be true, however solving gives either: $a=3/2$, $b=pm 1/4$ or $a=-3/2$, $b=pm 1/4$. Any of these solutions contradicts that $a$ and $b$ were integers.
edited Jan 21 at 23:52
idriskameni
757321
answered Jan 21 at 22:25
Jonty FreemanJonty Freeman
1
edited Jan 21 at 23:52
idriskameni
757321
edited Jan 21 at 23:52
idriskameni
757321
edited Jan 21 at 23:52
idriskameni
757321
757321
answered Jan 21 at 22:25
Jonty FreemanJonty Freeman
1
answered Jan 21 at 22:25
Jonty FreemanJonty Freeman
1
answered Jan 21 at 22:25
Jonty FreemanJonty Freeman
1
1
$begingroup$
$(a+2b)(a-2b) = a^2 - 4b^2 ne a^2 - 4b$.
$endgroup$
– fleablood
Jan 21 at 22:53
add a comment |
$begingroup$
$(a+2b)(a-2b) = a^2 - 4b^2 ne a^2 - 4b$.
$endgroup$
– fleablood
Jan 21 at 22:53
$begingroup$
$(a+2b)(a-2b) = a^2 - 4b^2 ne a^2 - 4b$.
$endgroup$
– fleablood
Jan 21 at 22:53
$begingroup$
$(a+2b)(a-2b) = a^2 - 4b^2 ne a^2 - 4b$.
$endgroup$
– fleablood
Jan 21 at 22:53
add a comment |
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Is it possile if a is odd? Is it possible if a is even.
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– fleablood
Oct 23 '16 at 0:11
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It doesnt say anything about it being odd or even in the question
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– Chance Gordon
Oct 23 '16 at 0:16
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So? That doesn't mean you can't say anything about it being odd or even. So, again. What happens if a is odd? What happens if a is even?
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– fleablood
Oct 23 '16 at 0:37
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Contradiction: Let $a^2 - 4b = 2$ then $a^2 = 4b + 2 = 2(2b + 1)$ so $a$ is even. Let $a = 2c$. $4c^2 = 2(2b + 1)$ so $2c^2 = 2b + 1$. Hmmmmm.....
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– fleablood
Oct 23 '16 at 0:40