Simple way of explaining the empty product
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I understand that 3 raised to the power of, say, 2 is 9 because 3 times 3 equals 9.
But is it possible to explain in same simple terms how 3 raised to the power of 0 is 1?
abstract-algebra
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up vote
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favorite
I understand that 3 raised to the power of, say, 2 is 9 because 3 times 3 equals 9.
But is it possible to explain in same simple terms how 3 raised to the power of 0 is 1?
abstract-algebra
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I understand that 3 raised to the power of, say, 2 is 9 because 3 times 3 equals 9.
But is it possible to explain in same simple terms how 3 raised to the power of 0 is 1?
abstract-algebra
I understand that 3 raised to the power of, say, 2 is 9 because 3 times 3 equals 9.
But is it possible to explain in same simple terms how 3 raised to the power of 0 is 1?
abstract-algebra
abstract-algebra
asked 29 mins ago
brilliant
1797
1797
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2 Answers
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up vote
4
down vote
accepted
$x^0$ doesn't mean anything before we define what it means. We can define it to mean anything, but we usually want to define things so that they make sense.
We start by defining $x^k$ as $underbracexcdot x cdots x_ktext times$ because it's a useful way of writing the product.
Using this definition, we can notice the following interesting property:
For any pair of integers $k_1, k_2$ such that $k_1 > k_2$ and any positive real number $xin mathbb R$ we have $$fracx^k_1x^k_2 = x^k_1-k_2$$
Now, we like this property. We want this property be true for other pairs of $k_1, k_2$. In order for this rule to also be true if $k_1=k_2$, we have to define $x^0=1$, so that's what we define it as. We do it because it's useful.
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up vote
5
down vote
$$3=3^1=3^1+0=3^1times3^0=3times3^0$$implying that $3^0=1$.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
$x^0$ doesn't mean anything before we define what it means. We can define it to mean anything, but we usually want to define things so that they make sense.
We start by defining $x^k$ as $underbracexcdot x cdots x_ktext times$ because it's a useful way of writing the product.
Using this definition, we can notice the following interesting property:
For any pair of integers $k_1, k_2$ such that $k_1 > k_2$ and any positive real number $xin mathbb R$ we have $$fracx^k_1x^k_2 = x^k_1-k_2$$
Now, we like this property. We want this property be true for other pairs of $k_1, k_2$. In order for this rule to also be true if $k_1=k_2$, we have to define $x^0=1$, so that's what we define it as. We do it because it's useful.
add a comment |Â
up vote
4
down vote
accepted
$x^0$ doesn't mean anything before we define what it means. We can define it to mean anything, but we usually want to define things so that they make sense.
We start by defining $x^k$ as $underbracexcdot x cdots x_ktext times$ because it's a useful way of writing the product.
Using this definition, we can notice the following interesting property:
For any pair of integers $k_1, k_2$ such that $k_1 > k_2$ and any positive real number $xin mathbb R$ we have $$fracx^k_1x^k_2 = x^k_1-k_2$$
Now, we like this property. We want this property be true for other pairs of $k_1, k_2$. In order for this rule to also be true if $k_1=k_2$, we have to define $x^0=1$, so that's what we define it as. We do it because it's useful.
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
$x^0$ doesn't mean anything before we define what it means. We can define it to mean anything, but we usually want to define things so that they make sense.
We start by defining $x^k$ as $underbracexcdot x cdots x_ktext times$ because it's a useful way of writing the product.
Using this definition, we can notice the following interesting property:
For any pair of integers $k_1, k_2$ such that $k_1 > k_2$ and any positive real number $xin mathbb R$ we have $$fracx^k_1x^k_2 = x^k_1-k_2$$
Now, we like this property. We want this property be true for other pairs of $k_1, k_2$. In order for this rule to also be true if $k_1=k_2$, we have to define $x^0=1$, so that's what we define it as. We do it because it's useful.
$x^0$ doesn't mean anything before we define what it means. We can define it to mean anything, but we usually want to define things so that they make sense.
We start by defining $x^k$ as $underbracexcdot x cdots x_ktext times$ because it's a useful way of writing the product.
Using this definition, we can notice the following interesting property:
For any pair of integers $k_1, k_2$ such that $k_1 > k_2$ and any positive real number $xin mathbb R$ we have $$fracx^k_1x^k_2 = x^k_1-k_2$$
Now, we like this property. We want this property be true for other pairs of $k_1, k_2$. In order for this rule to also be true if $k_1=k_2$, we have to define $x^0=1$, so that's what we define it as. We do it because it's useful.
answered 24 mins ago
5xum
84.3k386151
84.3k386151
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up vote
5
down vote
$$3=3^1=3^1+0=3^1times3^0=3times3^0$$implying that $3^0=1$.
add a comment |Â
up vote
5
down vote
$$3=3^1=3^1+0=3^1times3^0=3times3^0$$implying that $3^0=1$.
add a comment |Â
up vote
5
down vote
up vote
5
down vote
$$3=3^1=3^1+0=3^1times3^0=3times3^0$$implying that $3^0=1$.
$$3=3^1=3^1+0=3^1times3^0=3times3^0$$implying that $3^0=1$.
answered 24 mins ago
Vera
2,038414
2,038414
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