Number multiplied by itself does not give a square numberAre 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
The Digit Triangles
Why should universal income be universal?
Strong empirical falsification of quantum mechanics based on vacuum energy density?
Will number of steps recorded on FitBit/any fitness tracker add up distance in PokemonGo?
Is this toilet slogan correct usage of the English language?
What is the difference between lands and mana?
Short story about a deaf man, who cuts people tongues
How to explain what's wrong with this application of the chain rule?
What does Apple's new App Store requirement mean
Biological Blimps: Propulsion
How much theory knowledge is actually used while playing?
Can you use Vicious Mockery to win an argument or gain favours?
How would you translate "more" for use as an interface button?
Why is it that I can sometimes guess the next note?
Fantasy comedy romance novel, set in medieval times, involving a siege, a nun, and enchanted pickles
Taxes on Dividends in a Roth IRA
What features enable the Su-25 Frogfoot to operate with such a wide variety of fuels?
How to convince somebody that he is fit for something else, but not this job?
Why is the "ls" command showing permissions of files in a FAT32 partition?
Creating two special characters
How to get directions in deep space?
How can ping know if my host is down
Is there any evidence that Cleopatra and Caesarion considered fleeing to India to escape the Romans?
Were Persian-Median kings illiterate?
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 |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3148233%2fnumber-multiplied-by-itself-does-not-give-a-square-number%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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 |
$begingroup$
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3148233%2fnumber-multiplied-by-itself-does-not-give-a-square-number%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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