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Number multiplied by itself does not give a square number
Are square numbers also known as rectangular numbers, too?I call them squares. They called them arrays. What do they mean?When is $8x^2-4$ a square number?A property regarding complete/perfect squares.If the square of a natural number is odd then this number is odd.Can a multi-perfect number be a perfect square?What is the series of numbers, where each number is a triangular, square, and hexagonal number?Why is there a pattern to the last digits of square numbers?Is mean squared error equivalent to mean squared absolute errorThree-Digit Square Containing Numbers 1-9
$begingroup$
The answer to this is probably very simple but while working on a question I was surprised to discover than a number multiplied by itself does not give the same answer as the same number squared (in the case of negative numbers).
Examples:
$-1^2 = -1$
If I now multiply this number by itself $-1 * -1$ I get $1$
$-4^2 = -16$
$-4 * -4 = 16$
The same goes for all the other negative numbers I tried.
What changes when squaring a number and how do I understand what is going on? When I multiply a positive number by itself I get a square number, why does this not work with negative numbers?
square-numbers
asked Mar 14 at 16:47
SimonSimon
1104
$endgroup$
add a comment |
$begingroup$
The answer to this is probably very simple but while working on a question I was surprised to discover than a number multiplied by itself does not give the same answer as the same number squared (in the case of negative numbers).
Examples:
$-1^2 = -1$
If I now multiply this number by itself $-1 * -1$ I get $1$
$-4^2 = -16$
$-4 * -4 = 16$
The same goes for all the other negative numbers I tried.
What changes when squaring a number and how do I understand what is going on? When I multiply a positive number by itself I get a square number, why does this not work with negative numbers?
square-numbers
asked Mar 14 at 16:47
SimonSimon
1104
$endgroup$
4
$begingroup$
It's just operator precedence. $-x^2$ does not mean $(-x)^2$, it means $-(x^2)$.
$endgroup$
– Robert Israel
Mar 14 at 16:49
$begingroup$
What you've done above is not squaring a number. It is true (depending on how you interpret order of operations) that $-1^2=-1$. But what you are actually asking about is $(-1)^2$ which is exactly $1$. By definition, $x^2$ is the same thing as $xcdot x$.
$endgroup$
– Nick Peterson
Mar 14 at 16:50
$begingroup$
@NickPeterson So it's actually (-1)1^2
$endgroup$
– Simon
Mar 14 at 16:55
$begingroup$
@Robert Israel thank you I understand now
$endgroup$
– Simon
Mar 14 at 16:56
add a comment |
$begingroup$
The answer to this is probably very simple but while working on a question I was surprised to discover than a number multiplied by itself does not give the same answer as the same number squared (in the case of negative numbers).
Examples:
$-1^2 = -1$
If I now multiply this number by itself $-1 * -1$ I get $1$
$-4^2 = -16$
$-4 * -4 = 16$
The same goes for all the other negative numbers I tried.
What changes when squaring a number and how do I understand what is going on? When I multiply a positive number by itself I get a square number, why does this not work with negative numbers?
square-numbers
asked Mar 14 at 16:47
SimonSimon
1104
$endgroup$
The answer to this is probably very simple but while working on a question I was surprised to discover than a number multiplied by itself does not give the same answer as the same number squared (in the case of negative numbers).
Examples:
$-1^2 = -1$
If I now multiply this number by itself $-1 * -1$ I get $1$
$-4^2 = -16$
$-4 * -4 = 16$
The same goes for all the other negative numbers I tried.
What changes when squaring a number and how do I understand what is going on? When I multiply a positive number by itself I get a square number, why does this not work with negative numbers?
square-numbers
square-numbers
asked Mar 14 at 16:47
SimonSimon
1104
asked Mar 14 at 16:47
SimonSimon
1104
asked Mar 14 at 16:47
SimonSimon
1104
asked Mar 14 at 16:47
SimonSimon
1104
asked Mar 14 at 16:47
SimonSimon
1104
1104
4
$begingroup$
It's just operator precedence. $-x^2$ does not mean $(-x)^2$, it means $-(x^2)$.
$endgroup$
– Robert Israel
Mar 14 at 16:49
$begingroup$
What you've done above is not squaring a number. It is true (depending on how you interpret order of operations) that $-1^2=-1$. But what you are actually asking about is $(-1)^2$ which is exactly $1$. By definition, $x^2$ is the same thing as $xcdot x$.
$endgroup$
– Nick Peterson
Mar 14 at 16:50
$begingroup$
@NickPeterson So it's actually (-1)1^2
$endgroup$
– Simon
Mar 14 at 16:55
$begingroup$
@Robert Israel thank you I understand now
$endgroup$
– Simon
Mar 14 at 16:56
add a comment |
4
$begingroup$
It's just operator precedence. $-x^2$ does not mean $(-x)^2$, it means $-(x^2)$.
$endgroup$
– Robert Israel
Mar 14 at 16:49
$begingroup$
What you've done above is not squaring a number. It is true (depending on how you interpret order of operations) that $-1^2=-1$. But what you are actually asking about is $(-1)^2$ which is exactly $1$. By definition, $x^2$ is the same thing as $xcdot x$.
$endgroup$
– Nick Peterson
Mar 14 at 16:50
$begingroup$
@NickPeterson So it's actually (-1)1^2
$endgroup$
– Simon
Mar 14 at 16:55
$begingroup$
@Robert Israel thank you I understand now
$endgroup$
– Simon
Mar 14 at 16:56
4
4
$begingroup$
It's just operator precedence. $-x^2$ does not mean $(-x)^2$, it means $-(x^2)$.
$endgroup$
– Robert Israel
Mar 14 at 16:49
$begingroup$
It's just operator precedence. $-x^2$ does not mean $(-x)^2$, it means $-(x^2)$.
$endgroup$
– Robert Israel
Mar 14 at 16:49
$begingroup$
What you've done above is not squaring a number. It is true (depending on how you interpret order of operations) that $-1^2=-1$. But what you are actually asking about is $(-1)^2$ which is exactly $1$. By definition, $x^2$ is the same thing as $xcdot x$.
$endgroup$
– Nick Peterson
Mar 14 at 16:50
$begingroup$
What you've done above is not squaring a number. It is true (depending on how you interpret order of operations) that $-1^2=-1$. But what you are actually asking about is $(-1)^2$ which is exactly $1$. By definition, $x^2$ is the same thing as $xcdot x$.
$endgroup$
– Nick Peterson
Mar 14 at 16:50
$begingroup$
@NickPeterson So it's actually (-1)1^2
$endgroup$
– Simon
Mar 14 at 16:55
$begingroup$
@NickPeterson So it's actually (-1)1^2
$endgroup$
– Simon
Mar 14 at 16:55
$begingroup$
@Robert Israel thank you I understand now
$endgroup$
– Simon
Mar 14 at 16:56
$begingroup$
@Robert Israel thank you I understand now
$endgroup$
– Simon
Mar 14 at 16:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that the square of $-1$ is $$ (-1)^2 = (-1)times (-1) = 1 ne -1 .$$ You've mixed up operator precedence by putting $$ -(1)^2 = - (1) times (1) = - 1 . $$
In general $-(x)^2$ is $- (xcdot x)$, because exponentiation has a higher operator precedence than negation. You can read more about the order of operations here.
answered Mar 14 at 16:53
Dando18Dando18
4,73241235
$endgroup$
$begingroup$
I see I understand now. I'll accept your answer. Thank you, it was as obvious as I expected
$endgroup$
– Simon
Mar 14 at 16:59
add a comment |
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$begingroup$
Note that the square of $-1$ is $$ (-1)^2 = (-1)times (-1) = 1 ne -1 .$$ You've mixed up operator precedence by putting $$ -(1)^2 = - (1) times (1) = - 1 . $$
In general $-(x)^2$ is $- (xcdot x)$, because exponentiation has a higher operator precedence than negation. You can read more about the order of operations here.
answered Mar 14 at 16:53
Dando18Dando18
4,73241235
$endgroup$
$begingroup$
I see I understand now. I'll accept your answer. Thank you, it was as obvious as I expected
$endgroup$
– Simon
Mar 14 at 16:59
add a comment |
$begingroup$
Note that the square of $-1$ is $$ (-1)^2 = (-1)times (-1) = 1 ne -1 .$$ You've mixed up operator precedence by putting $$ -(1)^2 = - (1) times (1) = - 1 . $$
In general $-(x)^2$ is $- (xcdot x)$, because exponentiation has a higher operator precedence than negation. You can read more about the order of operations here.
answered Mar 14 at 16:53
Dando18Dando18
4,73241235
$endgroup$
$begingroup$
I see I understand now. I'll accept your answer. Thank you, it was as obvious as I expected
$endgroup$
– Simon
Mar 14 at 16:59
add a comment |
$begingroup$
Note that the square of $-1$ is $$ (-1)^2 = (-1)times (-1) = 1 ne -1 .$$ You've mixed up operator precedence by putting $$ -(1)^2 = - (1) times (1) = - 1 . $$
In general $-(x)^2$ is $- (xcdot x)$, because exponentiation has a higher operator precedence than negation. You can read more about the order of operations here.
answered Mar 14 at 16:53
Dando18Dando18
4,73241235
$endgroup$
Note that the square of $-1$ is $$ (-1)^2 = (-1)times (-1) = 1 ne -1 .$$ You've mixed up operator precedence by putting $$ -(1)^2 = - (1) times (1) = - 1 . $$
In general $-(x)^2$ is $- (xcdot x)$, because exponentiation has a higher operator precedence than negation. You can read more about the order of operations here.
answered Mar 14 at 16:53
Dando18Dando18
4,73241235
answered Mar 14 at 16:53
Dando18Dando18
4,73241235
answered Mar 14 at 16:53
Dando18Dando18
4,73241235
answered Mar 14 at 16:53
Dando18Dando18
4,73241235
4,73241235
$begingroup$
I see I understand now. I'll accept your answer. Thank you, it was as obvious as I expected
$endgroup$
– Simon
Mar 14 at 16:59
add a comment |
$begingroup$
I see I understand now. I'll accept your answer. Thank you, it was as obvious as I expected
$endgroup$
– Simon
Mar 14 at 16:59
$begingroup$
I see I understand now. I'll accept your answer. Thank you, it was as obvious as I expected
$endgroup$
– Simon
Mar 14 at 16:59
$begingroup$
I see I understand now. I'll accept your answer. Thank you, it was as obvious as I expected
$endgroup$
– Simon
Mar 14 at 16:59
add a comment |
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4
$begingroup$
It's just operator precedence. $-x^2$ does not mean $(-x)^2$, it means $-(x^2)$.
$endgroup$
– Robert Israel
Mar 14 at 16:49
$begingroup$
What you've done above is not squaring a number. It is true (depending on how you interpret order of operations) that $-1^2=-1$. But what you are actually asking about is $(-1)^2$ which is exactly $1$. By definition, $x^2$ is the same thing as $xcdot x$.
$endgroup$
– Nick Peterson
Mar 14 at 16:50
$begingroup$
@NickPeterson So it's actually (-1)1^2
$endgroup$
– Simon
Mar 14 at 16:55
$begingroup$
@Robert Israel thank you I understand now
$endgroup$
– Simon
Mar 14 at 16:56