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Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Why do we reverse inequality sign when dividing by negative number?Square root of a squared number changes sign, which to apply first?When solving inequalities, does $(x-9)$ count as a negative number?When do I use the 'plus-minus' sign when square rooting both sides of an equation? (example in main body).Restrictions on Factorial UsageWhen do I have to take the solution of a square root of a number with negative sign?Why does the negative sign leave when this expression is simplified?“Adding a negative” and other questions about the minus sign.On Changing The Direction Of The Inequality Sign By Dividing By A Negative Number?When do I apply the distributive property?
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I always had this confusion of when I need to apply the negative sign in the calculation.
I understand that $(-1)^2 = 1$ however why isn't $-1^2 = 1$?
algebra-precalculus recreational-mathematics
asked 4 hours ago
JohnJohnyPapaJohnJohnJohnyPapaJohn
626
$endgroup$
add a comment |
$begingroup$
I always had this confusion of when I need to apply the negative sign in the calculation.
I understand that $(-1)^2 = 1$ however why isn't $-1^2 = 1$?
algebra-precalculus recreational-mathematics
asked 4 hours ago
JohnJohnyPapaJohnJohnJohnyPapaJohn
626
$endgroup$
3
$begingroup$
because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
$endgroup$
– Luke
4 hours ago
$begingroup$
Though beware Excel and some similar cases, where=-1^2gives1but=0-1^2gives-1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first-as a unary operation taking precedence over exponentiation and the second-as a binary operation with exponentiation taking precedence over it
$endgroup$
– Henry
4 hours ago
$begingroup$
Just for an example, that's the same as writing $-1 times 1^2 = 1$, which probably is pretty clear that it's not true
$endgroup$
– MCMastery
2 hours ago
add a comment |
$begingroup$
I always had this confusion of when I need to apply the negative sign in the calculation.
I understand that $(-1)^2 = 1$ however why isn't $-1^2 = 1$?
algebra-precalculus recreational-mathematics
asked 4 hours ago
JohnJohnyPapaJohnJohnJohnyPapaJohn
626
$endgroup$
I always had this confusion of when I need to apply the negative sign in the calculation.
I understand that $(-1)^2 = 1$ however why isn't $-1^2 = 1$?
algebra-precalculus recreational-mathematics
algebra-precalculus recreational-mathematics
asked 4 hours ago
JohnJohnyPapaJohnJohnJohnyPapaJohn
626
asked 4 hours ago
JohnJohnyPapaJohnJohnJohnyPapaJohn
626
asked 4 hours ago
JohnJohnyPapaJohnJohnJohnyPapaJohn
626
asked 4 hours ago
JohnJohnyPapaJohnJohnJohnyPapaJohn
626
asked 4 hours ago
JohnJohnyPapaJohnJohnJohnyPapaJohn
626
626
3
$begingroup$
because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
$endgroup$
– Luke
4 hours ago
$begingroup$
Though beware Excel and some similar cases, where=-1^2gives1but=0-1^2gives-1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first-as a unary operation taking precedence over exponentiation and the second-as a binary operation with exponentiation taking precedence over it
$endgroup$
– Henry
4 hours ago
$begingroup$
Just for an example, that's the same as writing $-1 times 1^2 = 1$, which probably is pretty clear that it's not true
$endgroup$
– MCMastery
2 hours ago
add a comment |
3
$begingroup$
because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
$endgroup$
– Luke
4 hours ago
$begingroup$
Though beware Excel and some similar cases, where=-1^2gives1but=0-1^2gives-1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first-as a unary operation taking precedence over exponentiation and the second-as a binary operation with exponentiation taking precedence over it
$endgroup$
– Henry
4 hours ago
$begingroup$
Just for an example, that's the same as writing $-1 times 1^2 = 1$, which probably is pretty clear that it's not true
$endgroup$
– MCMastery
2 hours ago
3
3
$begingroup$
because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
$endgroup$
– Luke
4 hours ago
$begingroup$
because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
$endgroup$
– Luke
4 hours ago
$begingroup$
Though beware Excel and some similar cases, where
=-1^2 gives 1 but =0-1^2 gives -1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first - as a unary operation taking precedence over exponentiation and the second - as a binary operation with exponentiation taking precedence over it$endgroup$
– Henry
4 hours ago
$begingroup$
Though beware Excel and some similar cases, where
=-1^2 gives 1 but =0-1^2 gives -1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first - as a unary operation taking precedence over exponentiation and the second - as a binary operation with exponentiation taking precedence over it$endgroup$
– Henry
4 hours ago
$begingroup$
Just for an example, that's the same as writing $-1 times 1^2 = 1$, which probably is pretty clear that it's not true
$endgroup$
– MCMastery
2 hours ago
$begingroup$
Just for an example, that's the same as writing $-1 times 1^2 = 1$, which probably is pretty clear that it's not true
$endgroup$
– MCMastery
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
When we write $-x^2$, it means we square $x$ first, then take the negative of this. That is, $$-x^2 = -left(x^2right).$$ So $$-1^2 = -left(1^2right)=-1.$$ (And thus $-x^2$ means something different to $(-x)^2$.)
answered 4 hours ago
Minus One-TwelfthMinus One-Twelfth
3,628413
$endgroup$
add a comment |
$begingroup$
As it is already in the previous answers:
$(-x)^2neq-x^2$
To avoid confusion, it is better to use parentheses. $-x^2=-(x^2)$
answered 3 hours ago
user665960user665960
155
New contributor
user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
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2 Answers
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active
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2 Answers
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active
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$begingroup$
When we write $-x^2$, it means we square $x$ first, then take the negative of this. That is, $$-x^2 = -left(x^2right).$$ So $$-1^2 = -left(1^2right)=-1.$$ (And thus $-x^2$ means something different to $(-x)^2$.)
answered 4 hours ago
Minus One-TwelfthMinus One-Twelfth
3,628413
$endgroup$
add a comment |
$begingroup$
When we write $-x^2$, it means we square $x$ first, then take the negative of this. That is, $$-x^2 = -left(x^2right).$$ So $$-1^2 = -left(1^2right)=-1.$$ (And thus $-x^2$ means something different to $(-x)^2$.)
answered 4 hours ago
Minus One-TwelfthMinus One-Twelfth
3,628413
$endgroup$
add a comment |
$begingroup$
When we write $-x^2$, it means we square $x$ first, then take the negative of this. That is, $$-x^2 = -left(x^2right).$$ So $$-1^2 = -left(1^2right)=-1.$$ (And thus $-x^2$ means something different to $(-x)^2$.)
answered 4 hours ago
Minus One-TwelfthMinus One-Twelfth
3,628413
$endgroup$
When we write $-x^2$, it means we square $x$ first, then take the negative of this. That is, $$-x^2 = -left(x^2right).$$ So $$-1^2 = -left(1^2right)=-1.$$ (And thus $-x^2$ means something different to $(-x)^2$.)
answered 4 hours ago
Minus One-TwelfthMinus One-Twelfth
3,628413
answered 4 hours ago
Minus One-TwelfthMinus One-Twelfth
3,628413
answered 4 hours ago
Minus One-TwelfthMinus One-Twelfth
3,628413
answered 4 hours ago
Minus One-TwelfthMinus One-Twelfth
3,628413
3,628413
add a comment |
add a comment |
$begingroup$
As it is already in the previous answers:
$(-x)^2neq-x^2$
To avoid confusion, it is better to use parentheses. $-x^2=-(x^2)$
answered 3 hours ago
user665960user665960
155
New contributor
user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
As it is already in the previous answers:
$(-x)^2neq-x^2$
To avoid confusion, it is better to use parentheses. $-x^2=-(x^2)$
answered 3 hours ago
user665960user665960
155
New contributor
user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
As it is already in the previous answers:
$(-x)^2neq-x^2$
To avoid confusion, it is better to use parentheses. $-x^2=-(x^2)$
answered 3 hours ago
user665960user665960
155
New contributor
user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
As it is already in the previous answers:
$(-x)^2neq-x^2$
To avoid confusion, it is better to use parentheses. $-x^2=-(x^2)$
answered 3 hours ago
user665960user665960
155
New contributor
user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
user665960user665960
155
New contributor
user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
user665960user665960
155
answered 3 hours ago
user665960user665960
155
155
New contributor
user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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3
$begingroup$
because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
$endgroup$
– Luke
4 hours ago
$begingroup$
Though beware Excel and some similar cases, where
=-1^2gives1but=0-1^2gives-1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first-as a unary operation taking precedence over exponentiation and the second-as a binary operation with exponentiation taking precedence over it$endgroup$
– Henry
4 hours ago
$begingroup$
Just for an example, that's the same as writing $-1 times 1^2 = 1$, which probably is pretty clear that it's not true
$endgroup$
– MCMastery
2 hours ago