Hi

(1/3)^-2 = 9
what about if there is no parenthesis? and the exponent of -2 is on the numerator 1?
would the answer be 9 (implied parenthesis? or be 1/3?

thx

 

((1)^-2)/3 = 1/3

Not really very exciting --- eh?

 

I just wanted a second opinion since the answer sheet gave 9. It must have been a wrong answer then.

thx

 

Use the order of operations.

\(\frac{1^{-2}}{3} = \frac{\frac{1}{1^{2}}}{3} = \frac{1}{3}\)

 

Use the order of operations.

\(\frac{1^{2}}{3} = \frac{1*1}{3} = \frac{1}{3}\)

the exponent is -2 and not 2.

 

the exponent is -2 and not 2.

Makes no difference. A one to any power, positive or negative, is still....one!

 

is -1^3 the same as (-1)^3?

thx

 

the exponent is -2 and not 2.

It makes no difference at all-- 1 to the power of anything is still 1.

I did write it down in a rush though, so I've fixed it. 1^(-2) is the same as 1/(1^2)

 

my last question is about whether the exponent applies to the whole number including the minus which is in front of the number or it applies to the number without the minus if a parenthesis is not used?

 

my last question is about whether the exponent applies to the whole number including the minus which is in front of the number or it applies to the number without the minus if a parenthesis is not used?

Do you not know the order of operations?

PEMDAS:

Parentheses, Exponents, Multiplication & Division, Addition & Subtraction

That is the order you need to do things in. When it comes to multiplication and division, do it left to right. Same with addition and subtraction. Negation is just subtraction from 0.

 

then
-1^2 = -1
(-1)^2 = 1

Am I correct?

 

then
-1^2 = -1
(-1)^2 = 1

Am I correct?

Correct.

___________________

 

then
-1^2 = -1
(-1)^2 = 1

Am I correct?

Wait a minute!

Both expressions are identical and they are both equal to 1
-1 to any ODD power is -1
-1 to any EVEN power is +1
+1 to any power, ODD or EVEN, is +1

 

Wait a minute!

Both expressions are identical and they are both equal to 1
-1 to any ODD power is -1
-1 to any EVEN power is +1
+1 to any power, ODD or EVEN, is +1

But in the first case, according to PEMDAS, it is 1 squared and then the negative sign is applied. The second case removes any ambiguity, by explicitly stating it is negative 1 that is to be squared.

 

I was a bit surprised by the answer, as I was taught that: -1^2 =1 On searching, it appears current consensus is that : -1^2 = -1 , and the reason given on one of the "ask" sites is that the exponent "does not distribute to the minus."

I do not consider such "ask sites" very authoritative by a long shot, but that was all I could find.

It seems to me that one corollary of that reasoning is that there are no negative numbers per se. For example, consider the square root of 1. We (maybe only Papabravo and I) used to write the roots as +1 and -1 and could show that +1^2 = 1 and -1^2 = 1. In light of the new standard, it appears the correct way to write those roots would be 1 and (-1), as that is the only way that the squares would equal 1. That seems a bit clunky.

While I could not find "expert" answers for the -1^2 explanation, I did find several good sources that showed examples -- unfortunately none did the -1^2 problem. Of note, one considered an equation like:

\(\frac{3+5}{1+3}\)

The correct answer in that case is obtained by doing the additions before the division, "because the parentheses are implied." So, why can't the parentheses around -1 be "implied," as they were several years ago?

I have already learned the new rule and will adapt to it. Most important to me now is when did that change happen? Do you have a source for that?

John

 

Hello there,

(-1)^2=1
-(1^2)=-1
However:
-1^2 in ambiguous because it depends on the priority of the two symbols "-" (the minus sign) and "^" (the exponentiation sign).

Some calculators assign the priority of exponentiation above that of the unary operator "-", and others put the unary operator above that of exponentiation.
In algebra, we would probably take that last expression to mean:
(-1)*1^2
and then of course exponentiation takes precedence. But calculators and computer programs dont always do it that way.

For example:
y=3-1^2
In this case the minus sign is a binary operator, so there's no question that exponentiation takes precedence. But for:
y=-1^2
one calculator or program may compile this into:
y=(-1)^2
and another may do it as so:
y=-(1^2)

So you really have to know the context you are working under.

Some calculators also have a second minus sign that is used for the unary operator, and it may look the same or it may be raised up slightly from the binary version.

 

Under the supposition that exponentiation has higher precedence than unary minus then I stand corrected. One could also rewrite the expressions in RPN (I'm an HP, not a TI) and remove all ambiguity about what your intentions are.

1 2 ^ - = -1
1 - 2 ^ = 1

 

As I mentioned before, "-X" is simply a short way of saying "0 - X". According to the order of operations, the negative should be treated as subtraction, and thus -1^2 is the same as -(1^2), or -1.

That is how modern calculators work, for example, because negative is treated as subtraction in the standard order of operations. No ambiguity there.

 

As I mentioned before, "-X" is simply a short way of saying "0 - X". According to the order of operations, the negative should be treated as subtraction, and thus -1^2 is the same as -(1^2), or -1.

That is how modern calculators work, for example, because negative is treated as subtraction in the standard order of operations. No ambiguity there.

Except there is a difference between how binary subtraction and unary minus actually work and are actually implemented. In particular unary minus can be performed without doing subtraction and requires only a single operand on top of the evaluation stack.

In my TI-92 the binary minus followed by a constant actually computes:
ans(1) - const
that is the previous answer minus the constant.
The unary minus is just applied to the constant.

Clearly on this calculator they are different animals.

 

Except there is a difference between how binary subtraction and unary minus actually work and are actually implemented. In particular unary minus can be performed without doing subtraction and requires only a single operand on top of the evaluation stack.

In my TI-92 the binary minus followed by a constant actually computes:
ans(1) - const
that is the previous answer minus the constant.
The unary minus is just applied to the constant.

Clearly on this calculator they are different animals.

Perhaps, on the surface they may look different, but I believe the internal workings are the same. There is no "negative" in binary, so instead it uses two's complement and adds the value to zero. This is the same as subtracting from zero. On a software level, subtraction and negatives are handled the same way.

That is not to say binary negation works the same--that's just inverting the bits, but that's not what we're dealing with here.