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Understanding fractional exponents
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How do I understand fractional exponents like 1/2, 3/2, 1/5, etc?
Like $y$ raised to a power 2 is $y.y$ and when the power is increased or decreased by one, it means we multiply the number by another $y$ or divide it by $y$ respectively.
But how do I think about fractional exponential powers by this same logic?
Thank you.
exponential-function
asked Jul 22 at 16:23
Ram Keswani
32112
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up vote
0
down vote
favorite
How do I understand fractional exponents like 1/2, 3/2, 1/5, etc?
Like $y$ raised to a power 2 is $y.y$ and when the power is increased or decreased by one, it means we multiply the number by another $y$ or divide it by $y$ respectively.
But how do I think about fractional exponential powers by this same logic?
Thank you.
exponential-function
asked Jul 22 at 16:23
Ram Keswani
32112
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How do I understand fractional exponents like 1/2, 3/2, 1/5, etc?
Like $y$ raised to a power 2 is $y.y$ and when the power is increased or decreased by one, it means we multiply the number by another $y$ or divide it by $y$ respectively.
But how do I think about fractional exponential powers by this same logic?
Thank you.
exponential-function
asked Jul 22 at 16:23
Ram Keswani
32112
How do I understand fractional exponents like 1/2, 3/2, 1/5, etc?
Like $y$ raised to a power 2 is $y.y$ and when the power is increased or decreased by one, it means we multiply the number by another $y$ or divide it by $y$ respectively.
But how do I think about fractional exponential powers by this same logic?
Thank you.
exponential-function
asked Jul 22 at 16:23
Ram Keswani
32112
asked Jul 22 at 16:23
Ram Keswani
32112
asked Jul 22 at 16:23
Ram Keswani
32112
asked Jul 22 at 16:23
Ram Keswani
32112
32112
add a comment |Â
add a comment |Â
4 Answers
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Notice that, for example, if $x^frac12=a$, then by squaring both sides we get $x=(x^frac12)^2=a^2$.
In other words $sqrtx=a$.
Following the same logic you can get that for every $n in mathbbN$, $x^frac1n = sqrt[n]x$
After you accepted that, by exponentiation rules, you know that for every $p,q in mathbbN$:
$$
x^fracpq = x^p cdot frac1q = (x^p)^frac1q = sqrt[q]x^p
$$
So in general, raising to a rational number $fracpq$ means to take the $p$-th power and the $q$-th root.
And, of course, it also generalize to negative rational numbers.
answered Jul 22 at 16:33
GuySa
402313
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"how do I think about fractional exponential powers by this same logic?"
Carefully.
You have to expand the definition.
We notice that $b^n*b^m = underbraceb*...*b_nunderbraceb*...*b_m=underbraceb*...*b= b^n+m$ for positive integers $n$ and $m$ and we take $b^n+m = b^nb^m$ as a rule.
Then we think. Well what that means $b^0*b^m = b^0+m = b^m$. That would mean $b^0 =1$, doesn't it? Well, in one sense it does not make sense that we can't mulitple $b$ times itself zero times. That's meaningless.
But that is no longer what the definition of $b^n$ means. It means that for $nin mathbb Z$ and $n ge 2$ but... for $nin mathbb Z$ and $n le 1$ we are going to have it mean.... whatever it has to be in order for the rule $b^n+m = b^n*b^m$ to work.
That means: $b^1 = b$ and $b^0 = 0$ and for a negative integer $-n < 0$ then $b^-n = frac 1b^n$ (becuase we must have $b^-n*b^n = b^-n+n = b^0 = 1$.
Okay, but we also have the rule $(b^n)^m = underbraceb^n*...*b^n_m = b^underbracen+....+n_m = b^nm$.
So what does that make $b^frac 1m$ have to be?
Well, $(b^frac 1m)^m = b^frac 1m*m = b^1 = b$. So that means $b^frac 1m$ has to be $sqrt[m]b$, right?
And that's that. $b^frac mn = (sqrt[n]b)^m$.
That's it.
....
Except we have to be careful. $x^2 = 4$ has two possible answer $-2,2$ so which will it be. ANd $x^2 = -4$ has none.
So we add a caveat rule: we only feel safe talking about $b^frac mn$ if $b ge 0$ if $n$ is even and in that case the value is positive. We can talk about $(-4)^frac mn$ if $n$ is odd (in which case the value is negative) but for most of the time, we should avoid it. Many mathematicians will simply say $b^q;qin mathbb Q$ is only defined if $bge 0$.
And for $x in mathbb R$, we say $b^x$ is only defined fro $bge 0$. Unless we are doing complex analysis where we accept the square roots of negative numbers. In that case, things get very different. But it's still a matter of defining things by what the must be in order for the rules to work.
answered Jul 22 at 16:43
fleablood
60.4k22575
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The exponents follow the rule
$$(a^b)^c=a^bc$$
meaning that the power of a power is the power to the product of the exponents.
Then assume a rational power $p/q$. We write
$$a^p/q=b$$ and raise both members to the $q^th$:
$$a^(p/q)q=a^p=b^q.$$
So $b$ is the number such that when raised to the $q^th$ power yields $a^p$. In other words, it is the $q^th$ root of $a^p$.
If you didn't learn the roots yet, understand that this is the "inverse" operation of exponentiation. For instance,
$$3^5=243iff243^1/5=sqrt[5]243=3.$$
Very often, roots are irrational numbers,
$$1.24573093cdots^5=3iff 3^1/5=sqrt[5]3=1.24573093cdots$$
answered Jul 22 at 16:34
Yves Daoust
111k665203
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Let us start with $1/2$
If I want to find $16^1/2$, I expect to find a number whose square is $16.$
That is if $$x= 16^1/2$$ we like to have $x^2=16$
Well we have two such real numbers, namely $4$ and $-4$
We denote them by $ sqrt 16 =4$ and $- sqrt 16 =-4$
Sometimes you do not get a real solution at all, for example $(-16)^1/2$ is not a real number, because when you square a real number the answer is always non-negative.
Sometimes you get only one solution for example $27^1/3=3$ because 3^3=27 and no other real number satisfies $x^3=27.$
Once you find $a^1/n$ then $a^m/n$ is found by $(a^1/n)^m$
For example $$27^2/3 = (27^1/3)^2 = 3^2 =9 $$
You may practice with some fractional powers and check your answers with a calculator to build confidence.
answered Jul 22 at 17:09
Mohammad Riazi-Kermani
27.5k41852
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Notice that, for example, if $x^frac12=a$, then by squaring both sides we get $x=(x^frac12)^2=a^2$.
In other words $sqrtx=a$.
Following the same logic you can get that for every $n in mathbbN$, $x^frac1n = sqrt[n]x$
After you accepted that, by exponentiation rules, you know that for every $p,q in mathbbN$:
$$
x^fracpq = x^p cdot frac1q = (x^p)^frac1q = sqrt[q]x^p
$$
So in general, raising to a rational number $fracpq$ means to take the $p$-th power and the $q$-th root.
And, of course, it also generalize to negative rational numbers.
answered Jul 22 at 16:33
GuySa
402313
add a comment |Â
up vote
0
down vote
Notice that, for example, if $x^frac12=a$, then by squaring both sides we get $x=(x^frac12)^2=a^2$.
In other words $sqrtx=a$.
Following the same logic you can get that for every $n in mathbbN$, $x^frac1n = sqrt[n]x$
After you accepted that, by exponentiation rules, you know that for every $p,q in mathbbN$:
$$
x^fracpq = x^p cdot frac1q = (x^p)^frac1q = sqrt[q]x^p
$$
So in general, raising to a rational number $fracpq$ means to take the $p$-th power and the $q$-th root.
And, of course, it also generalize to negative rational numbers.
answered Jul 22 at 16:33
GuySa
402313
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Notice that, for example, if $x^frac12=a$, then by squaring both sides we get $x=(x^frac12)^2=a^2$.
In other words $sqrtx=a$.
Following the same logic you can get that for every $n in mathbbN$, $x^frac1n = sqrt[n]x$
After you accepted that, by exponentiation rules, you know that for every $p,q in mathbbN$:
$$
x^fracpq = x^p cdot frac1q = (x^p)^frac1q = sqrt[q]x^p
$$
So in general, raising to a rational number $fracpq$ means to take the $p$-th power and the $q$-th root.
And, of course, it also generalize to negative rational numbers.
answered Jul 22 at 16:33
GuySa
402313
Notice that, for example, if $x^frac12=a$, then by squaring both sides we get $x=(x^frac12)^2=a^2$.
In other words $sqrtx=a$.
Following the same logic you can get that for every $n in mathbbN$, $x^frac1n = sqrt[n]x$
After you accepted that, by exponentiation rules, you know that for every $p,q in mathbbN$:
$$
x^fracpq = x^p cdot frac1q = (x^p)^frac1q = sqrt[q]x^p
$$
So in general, raising to a rational number $fracpq$ means to take the $p$-th power and the $q$-th root.
And, of course, it also generalize to negative rational numbers.
answered Jul 22 at 16:33
GuySa
402313
answered Jul 22 at 16:33
GuySa
402313
answered Jul 22 at 16:33
GuySa
402313
answered Jul 22 at 16:33
GuySa
402313
402313
add a comment |Â
add a comment |Â
up vote
0
down vote
"how do I think about fractional exponential powers by this same logic?"
Carefully.
You have to expand the definition.
We notice that $b^n*b^m = underbraceb*...*b_nunderbraceb*...*b_m=underbraceb*...*b= b^n+m$ for positive integers $n$ and $m$ and we take $b^n+m = b^nb^m$ as a rule.
Then we think. Well what that means $b^0*b^m = b^0+m = b^m$. That would mean $b^0 =1$, doesn't it? Well, in one sense it does not make sense that we can't mulitple $b$ times itself zero times. That's meaningless.
But that is no longer what the definition of $b^n$ means. It means that for $nin mathbb Z$ and $n ge 2$ but... for $nin mathbb Z$ and $n le 1$ we are going to have it mean.... whatever it has to be in order for the rule $b^n+m = b^n*b^m$ to work.
That means: $b^1 = b$ and $b^0 = 0$ and for a negative integer $-n < 0$ then $b^-n = frac 1b^n$ (becuase we must have $b^-n*b^n = b^-n+n = b^0 = 1$.
Okay, but we also have the rule $(b^n)^m = underbraceb^n*...*b^n_m = b^underbracen+....+n_m = b^nm$.
So what does that make $b^frac 1m$ have to be?
Well, $(b^frac 1m)^m = b^frac 1m*m = b^1 = b$. So that means $b^frac 1m$ has to be $sqrt[m]b$, right?
And that's that. $b^frac mn = (sqrt[n]b)^m$.
That's it.
....
Except we have to be careful. $x^2 = 4$ has two possible answer $-2,2$ so which will it be. ANd $x^2 = -4$ has none.
So we add a caveat rule: we only feel safe talking about $b^frac mn$ if $b ge 0$ if $n$ is even and in that case the value is positive. We can talk about $(-4)^frac mn$ if $n$ is odd (in which case the value is negative) but for most of the time, we should avoid it. Many mathematicians will simply say $b^q;qin mathbb Q$ is only defined if $bge 0$.
And for $x in mathbb R$, we say $b^x$ is only defined fro $bge 0$. Unless we are doing complex analysis where we accept the square roots of negative numbers. In that case, things get very different. But it's still a matter of defining things by what the must be in order for the rules to work.
answered Jul 22 at 16:43
fleablood
60.4k22575
add a comment |Â
up vote
0
down vote
"how do I think about fractional exponential powers by this same logic?"
Carefully.
You have to expand the definition.
We notice that $b^n*b^m = underbraceb*...*b_nunderbraceb*...*b_m=underbraceb*...*b= b^n+m$ for positive integers $n$ and $m$ and we take $b^n+m = b^nb^m$ as a rule.
Then we think. Well what that means $b^0*b^m = b^0+m = b^m$. That would mean $b^0 =1$, doesn't it? Well, in one sense it does not make sense that we can't mulitple $b$ times itself zero times. That's meaningless.
But that is no longer what the definition of $b^n$ means. It means that for $nin mathbb Z$ and $n ge 2$ but... for $nin mathbb Z$ and $n le 1$ we are going to have it mean.... whatever it has to be in order for the rule $b^n+m = b^n*b^m$ to work.
That means: $b^1 = b$ and $b^0 = 0$ and for a negative integer $-n < 0$ then $b^-n = frac 1b^n$ (becuase we must have $b^-n*b^n = b^-n+n = b^0 = 1$.
Okay, but we also have the rule $(b^n)^m = underbraceb^n*...*b^n_m = b^underbracen+....+n_m = b^nm$.
So what does that make $b^frac 1m$ have to be?
Well, $(b^frac 1m)^m = b^frac 1m*m = b^1 = b$. So that means $b^frac 1m$ has to be $sqrt[m]b$, right?
And that's that. $b^frac mn = (sqrt[n]b)^m$.
That's it.
....
Except we have to be careful. $x^2 = 4$ has two possible answer $-2,2$ so which will it be. ANd $x^2 = -4$ has none.
So we add a caveat rule: we only feel safe talking about $b^frac mn$ if $b ge 0$ if $n$ is even and in that case the value is positive. We can talk about $(-4)^frac mn$ if $n$ is odd (in which case the value is negative) but for most of the time, we should avoid it. Many mathematicians will simply say $b^q;qin mathbb Q$ is only defined if $bge 0$.
And for $x in mathbb R$, we say $b^x$ is only defined fro $bge 0$. Unless we are doing complex analysis where we accept the square roots of negative numbers. In that case, things get very different. But it's still a matter of defining things by what the must be in order for the rules to work.
answered Jul 22 at 16:43
fleablood
60.4k22575
add a comment |Â
up vote
0
down vote
up vote
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down vote
"how do I think about fractional exponential powers by this same logic?"
Carefully.
You have to expand the definition.
We notice that $b^n*b^m = underbraceb*...*b_nunderbraceb*...*b_m=underbraceb*...*b= b^n+m$ for positive integers $n$ and $m$ and we take $b^n+m = b^nb^m$ as a rule.
Then we think. Well what that means $b^0*b^m = b^0+m = b^m$. That would mean $b^0 =1$, doesn't it? Well, in one sense it does not make sense that we can't mulitple $b$ times itself zero times. That's meaningless.
But that is no longer what the definition of $b^n$ means. It means that for $nin mathbb Z$ and $n ge 2$ but... for $nin mathbb Z$ and $n le 1$ we are going to have it mean.... whatever it has to be in order for the rule $b^n+m = b^n*b^m$ to work.
That means: $b^1 = b$ and $b^0 = 0$ and for a negative integer $-n < 0$ then $b^-n = frac 1b^n$ (becuase we must have $b^-n*b^n = b^-n+n = b^0 = 1$.
Okay, but we also have the rule $(b^n)^m = underbraceb^n*...*b^n_m = b^underbracen+....+n_m = b^nm$.
So what does that make $b^frac 1m$ have to be?
Well, $(b^frac 1m)^m = b^frac 1m*m = b^1 = b$. So that means $b^frac 1m$ has to be $sqrt[m]b$, right?
And that's that. $b^frac mn = (sqrt[n]b)^m$.
That's it.
....
Except we have to be careful. $x^2 = 4$ has two possible answer $-2,2$ so which will it be. ANd $x^2 = -4$ has none.
So we add a caveat rule: we only feel safe talking about $b^frac mn$ if $b ge 0$ if $n$ is even and in that case the value is positive. We can talk about $(-4)^frac mn$ if $n$ is odd (in which case the value is negative) but for most of the time, we should avoid it. Many mathematicians will simply say $b^q;qin mathbb Q$ is only defined if $bge 0$.
And for $x in mathbb R$, we say $b^x$ is only defined fro $bge 0$. Unless we are doing complex analysis where we accept the square roots of negative numbers. In that case, things get very different. But it's still a matter of defining things by what the must be in order for the rules to work.
answered Jul 22 at 16:43
fleablood
60.4k22575
"how do I think about fractional exponential powers by this same logic?"
Carefully.
You have to expand the definition.
We notice that $b^n*b^m = underbraceb*...*b_nunderbraceb*...*b_m=underbraceb*...*b= b^n+m$ for positive integers $n$ and $m$ and we take $b^n+m = b^nb^m$ as a rule.
Then we think. Well what that means $b^0*b^m = b^0+m = b^m$. That would mean $b^0 =1$, doesn't it? Well, in one sense it does not make sense that we can't mulitple $b$ times itself zero times. That's meaningless.
But that is no longer what the definition of $b^n$ means. It means that for $nin mathbb Z$ and $n ge 2$ but... for $nin mathbb Z$ and $n le 1$ we are going to have it mean.... whatever it has to be in order for the rule $b^n+m = b^n*b^m$ to work.
That means: $b^1 = b$ and $b^0 = 0$ and for a negative integer $-n < 0$ then $b^-n = frac 1b^n$ (becuase we must have $b^-n*b^n = b^-n+n = b^0 = 1$.
Okay, but we also have the rule $(b^n)^m = underbraceb^n*...*b^n_m = b^underbracen+....+n_m = b^nm$.
So what does that make $b^frac 1m$ have to be?
Well, $(b^frac 1m)^m = b^frac 1m*m = b^1 = b$. So that means $b^frac 1m$ has to be $sqrt[m]b$, right?
And that's that. $b^frac mn = (sqrt[n]b)^m$.
That's it.
....
Except we have to be careful. $x^2 = 4$ has two possible answer $-2,2$ so which will it be. ANd $x^2 = -4$ has none.
So we add a caveat rule: we only feel safe talking about $b^frac mn$ if $b ge 0$ if $n$ is even and in that case the value is positive. We can talk about $(-4)^frac mn$ if $n$ is odd (in which case the value is negative) but for most of the time, we should avoid it. Many mathematicians will simply say $b^q;qin mathbb Q$ is only defined if $bge 0$.
And for $x in mathbb R$, we say $b^x$ is only defined fro $bge 0$. Unless we are doing complex analysis where we accept the square roots of negative numbers. In that case, things get very different. But it's still a matter of defining things by what the must be in order for the rules to work.
answered Jul 22 at 16:43
fleablood
60.4k22575
edited Jul 22 at 16:50
answered Jul 22 at 16:43
fleablood
60.4k22575
answered Jul 22 at 16:43
fleablood
60.4k22575
answered Jul 22 at 16:43
fleablood
60.4k22575
60.4k22575
add a comment |Â
add a comment |Â
up vote
0
down vote
The exponents follow the rule
$$(a^b)^c=a^bc$$
meaning that the power of a power is the power to the product of the exponents.
Then assume a rational power $p/q$. We write
$$a^p/q=b$$ and raise both members to the $q^th$:
$$a^(p/q)q=a^p=b^q.$$
So $b$ is the number such that when raised to the $q^th$ power yields $a^p$. In other words, it is the $q^th$ root of $a^p$.
If you didn't learn the roots yet, understand that this is the "inverse" operation of exponentiation. For instance,
$$3^5=243iff243^1/5=sqrt[5]243=3.$$
Very often, roots are irrational numbers,
$$1.24573093cdots^5=3iff 3^1/5=sqrt[5]3=1.24573093cdots$$
answered Jul 22 at 16:34
Yves Daoust
111k665203
add a comment |Â
up vote
0
down vote
The exponents follow the rule
$$(a^b)^c=a^bc$$
meaning that the power of a power is the power to the product of the exponents.
Then assume a rational power $p/q$. We write
$$a^p/q=b$$ and raise both members to the $q^th$:
$$a^(p/q)q=a^p=b^q.$$
So $b$ is the number such that when raised to the $q^th$ power yields $a^p$. In other words, it is the $q^th$ root of $a^p$.
If you didn't learn the roots yet, understand that this is the "inverse" operation of exponentiation. For instance,
$$3^5=243iff243^1/5=sqrt[5]243=3.$$
Very often, roots are irrational numbers,
$$1.24573093cdots^5=3iff 3^1/5=sqrt[5]3=1.24573093cdots$$
answered Jul 22 at 16:34
Yves Daoust
111k665203
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The exponents follow the rule
$$(a^b)^c=a^bc$$
meaning that the power of a power is the power to the product of the exponents.
Then assume a rational power $p/q$. We write
$$a^p/q=b$$ and raise both members to the $q^th$:
$$a^(p/q)q=a^p=b^q.$$
So $b$ is the number such that when raised to the $q^th$ power yields $a^p$. In other words, it is the $q^th$ root of $a^p$.
If you didn't learn the roots yet, understand that this is the "inverse" operation of exponentiation. For instance,
$$3^5=243iff243^1/5=sqrt[5]243=3.$$
Very often, roots are irrational numbers,
$$1.24573093cdots^5=3iff 3^1/5=sqrt[5]3=1.24573093cdots$$
answered Jul 22 at 16:34
Yves Daoust
111k665203
The exponents follow the rule
$$(a^b)^c=a^bc$$
meaning that the power of a power is the power to the product of the exponents.
Then assume a rational power $p/q$. We write
$$a^p/q=b$$ and raise both members to the $q^th$:
$$a^(p/q)q=a^p=b^q.$$
So $b$ is the number such that when raised to the $q^th$ power yields $a^p$. In other words, it is the $q^th$ root of $a^p$.
If you didn't learn the roots yet, understand that this is the "inverse" operation of exponentiation. For instance,
$$3^5=243iff243^1/5=sqrt[5]243=3.$$
Very often, roots are irrational numbers,
$$1.24573093cdots^5=3iff 3^1/5=sqrt[5]3=1.24573093cdots$$
answered Jul 22 at 16:34
Yves Daoust
111k665203
edited Jul 22 at 17:08
answered Jul 22 at 16:34
Yves Daoust
111k665203
answered Jul 22 at 16:34
Yves Daoust
111k665203
answered Jul 22 at 16:34
Yves Daoust
111k665203
111k665203
add a comment |Â
add a comment |Â
up vote
0
down vote
Let us start with $1/2$
If I want to find $16^1/2$, I expect to find a number whose square is $16.$
That is if $$x= 16^1/2$$ we like to have $x^2=16$
Well we have two such real numbers, namely $4$ and $-4$
We denote them by $ sqrt 16 =4$ and $- sqrt 16 =-4$
Sometimes you do not get a real solution at all, for example $(-16)^1/2$ is not a real number, because when you square a real number the answer is always non-negative.
Sometimes you get only one solution for example $27^1/3=3$ because 3^3=27 and no other real number satisfies $x^3=27.$
Once you find $a^1/n$ then $a^m/n$ is found by $(a^1/n)^m$
For example $$27^2/3 = (27^1/3)^2 = 3^2 =9 $$
You may practice with some fractional powers and check your answers with a calculator to build confidence.
answered Jul 22 at 17:09
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
0
down vote
Let us start with $1/2$
If I want to find $16^1/2$, I expect to find a number whose square is $16.$
That is if $$x= 16^1/2$$ we like to have $x^2=16$
Well we have two such real numbers, namely $4$ and $-4$
We denote them by $ sqrt 16 =4$ and $- sqrt 16 =-4$
Sometimes you do not get a real solution at all, for example $(-16)^1/2$ is not a real number, because when you square a real number the answer is always non-negative.
Sometimes you get only one solution for example $27^1/3=3$ because 3^3=27 and no other real number satisfies $x^3=27.$
Once you find $a^1/n$ then $a^m/n$ is found by $(a^1/n)^m$
For example $$27^2/3 = (27^1/3)^2 = 3^2 =9 $$
You may practice with some fractional powers and check your answers with a calculator to build confidence.
answered Jul 22 at 17:09
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let us start with $1/2$
If I want to find $16^1/2$, I expect to find a number whose square is $16.$
That is if $$x= 16^1/2$$ we like to have $x^2=16$
Well we have two such real numbers, namely $4$ and $-4$
We denote them by $ sqrt 16 =4$ and $- sqrt 16 =-4$
Sometimes you do not get a real solution at all, for example $(-16)^1/2$ is not a real number, because when you square a real number the answer is always non-negative.
Sometimes you get only one solution for example $27^1/3=3$ because 3^3=27 and no other real number satisfies $x^3=27.$
Once you find $a^1/n$ then $a^m/n$ is found by $(a^1/n)^m$
For example $$27^2/3 = (27^1/3)^2 = 3^2 =9 $$
You may practice with some fractional powers and check your answers with a calculator to build confidence.
answered Jul 22 at 17:09
Mohammad Riazi-Kermani
27.5k41852
Let us start with $1/2$
If I want to find $16^1/2$, I expect to find a number whose square is $16.$
That is if $$x= 16^1/2$$ we like to have $x^2=16$
Well we have two such real numbers, namely $4$ and $-4$
We denote them by $ sqrt 16 =4$ and $- sqrt 16 =-4$
Sometimes you do not get a real solution at all, for example $(-16)^1/2$ is not a real number, because when you square a real number the answer is always non-negative.
Sometimes you get only one solution for example $27^1/3=3$ because 3^3=27 and no other real number satisfies $x^3=27.$
Once you find $a^1/n$ then $a^m/n$ is found by $(a^1/n)^m$
For example $$27^2/3 = (27^1/3)^2 = 3^2 =9 $$
You may practice with some fractional powers and check your answers with a calculator to build confidence.
answered Jul 22 at 17:09
Mohammad Riazi-Kermani
27.5k41852
answered Jul 22 at 17:09
Mohammad Riazi-Kermani
27.5k41852
answered Jul 22 at 17:09
Mohammad Riazi-Kermani
27.5k41852
answered Jul 22 at 17:09
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
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