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0 to the power of any number

0

$begingroup$

I have proof that $0^n$ = undefined.

Since,
$2^5 = 32$,
$2^4 = 16$,
$2^5/2 = 32/2 = 16 = 2^4$.

Similarly if $0^n = 0$.

Then,
$0^n-1 = 0$
$0^0/0 = 0/0 = 0^n-1$.

But $0/0$ is undefined.
Therefore $0^n = 0$.

But calculators give the result of $0^n$ as $0$. Can you explain where I am going wrong.

proof-verification exponentiation

share|cite|improve this question

edited Mar 27 at 19:01

Somos

15.1k11437

asked Mar 27 at 16:43

Anirudh PanguluriAnirudh Panguluri

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $0^n$ see the answers below. But let me point out that your "proof" for $2^5=2^4$ also isn't correct.
  $endgroup$
  – James
  Mar 27 at 16:53
  

  
  
  
  

add a comment | 

0

$begingroup$

I have proof that $0^n$ = undefined.

Since,
$2^5 = 32$,
$2^4 = 16$,
$2^5/2 = 32/2 = 16 = 2^4$.

Similarly if $0^n = 0$.

Then,
$0^n-1 = 0$
$0^0/0 = 0/0 = 0^n-1$.

But $0/0$ is undefined.
Therefore $0^n = 0$.

But calculators give the result of $0^n$ as $0$. Can you explain where I am going wrong.

proof-verification exponentiation

share|cite|improve this question

edited Mar 27 at 19:01

Somos

15.1k11437

asked Mar 27 at 16:43

Anirudh PanguluriAnirudh Panguluri

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $0^n$ see the answers below. But let me point out that your "proof" for $2^5=2^4$ also isn't correct.
  $endgroup$
  – James
  Mar 27 at 16:53
  

  
  
  
  

add a comment | 

$begingroup$

I have proof that $0^n$ = undefined.

Since,
$2^5 = 32$,
$2^4 = 16$,
$2^5/2 = 32/2 = 16 = 2^4$.

Similarly if $0^n = 0$.

Then,
$0^n-1 = 0$
$0^0/0 = 0/0 = 0^n-1$.

But $0/0$ is undefined.
Therefore $0^n = 0$.

But calculators give the result of $0^n$ as $0$. Can you explain where I am going wrong.

proof-verification exponentiation

share|cite|improve this question

edited Mar 27 at 19:01

Somos

15.1k11437

asked Mar 27 at 16:43

Anirudh PanguluriAnirudh Panguluri

1

$endgroup$

share|cite|improve this question

edited Mar 27 at 19:01

Somos

15.1k11437

asked Mar 27 at 16:43

Anirudh PanguluriAnirudh Panguluri

1

share|cite|improve this question

edited Mar 27 at 19:01

Somos

15.1k11437

asked Mar 27 at 16:43

Anirudh PanguluriAnirudh Panguluri

1

edited Mar 27 at 19:01

Somos

15.1k11437

edited Mar 27 at 19:01

Somos

15.1k11437

asked Mar 27 at 16:43

Anirudh PanguluriAnirudh Panguluri

1

asked Mar 27 at 16:43

Anirudh PanguluriAnirudh Panguluri

1

2 Answers
2

active

oldest

votes

4

$begingroup$

The rule $x^n-1=x^n/x$ requires that you divide by $x$, which you can't when $x=0$. The rule comes from
$$
x^n=x^n-1 x,
$$
and then dividing by $x$. But if $x$ is zero you cannot divide, and so the rule does not apply.

share|cite|improve this answer

answered Mar 27 at 16:47

Martin ArgeramiMartin Argerami

130k1184185

$endgroup$

add a comment | 

0

$begingroup$

To calculate $0^n-1$, you divided by $0$, which is undefined. Notice that $0^n=0cdots0=0$ (with $n$ factors); there is no division involved here.

share|cite|improve this answer

answered Mar 27 at 16:47

st.mathst.math

1,268115

$endgroup$

add a comment | 

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4

$begingroup$

The rule $x^n-1=x^n/x$ requires that you divide by $x$, which you can't when $x=0$. The rule comes from
$$
x^n=x^n-1 x,
$$
and then dividing by $x$. But if $x$ is zero you cannot divide, and so the rule does not apply.

share|cite|improve this answer

answered Mar 27 at 16:47

Martin ArgeramiMartin Argerami

130k1184185

$endgroup$

add a comment | 

4

$begingroup$

The rule $x^n-1=x^n/x$ requires that you divide by $x$, which you can't when $x=0$. The rule comes from
$$
x^n=x^n-1 x,
$$
and then dividing by $x$. But if $x$ is zero you cannot divide, and so the rule does not apply.

share|cite|improve this answer

answered Mar 27 at 16:47

Martin ArgeramiMartin Argerami

130k1184185

$endgroup$

add a comment | 

$begingroup$

The rule $x^n-1=x^n/x$ requires that you divide by $x$, which you can't when $x=0$. The rule comes from
$$
x^n=x^n-1 x,
$$
and then dividing by $x$. But if $x$ is zero you cannot divide, and so the rule does not apply.

share|cite|improve this answer

answered Mar 27 at 16:47

Martin ArgeramiMartin Argerami

130k1184185

$endgroup$

share|cite|improve this answer

answered Mar 27 at 16:47

Martin ArgeramiMartin Argerami

130k1184185

answered Mar 27 at 16:47

Martin ArgeramiMartin Argerami

130k1184185

answered Mar 27 at 16:47

Martin ArgeramiMartin Argerami

130k1184185

0

$begingroup$

To calculate $0^n-1$, you divided by $0$, which is undefined. Notice that $0^n=0cdots0=0$ (with $n$ factors); there is no division involved here.

share|cite|improve this answer

answered Mar 27 at 16:47

st.mathst.math

1,268115

$endgroup$

add a comment | 

0

$begingroup$

To calculate $0^n-1$, you divided by $0$, which is undefined. Notice that $0^n=0cdots0=0$ (with $n$ factors); there is no division involved here.

share|cite|improve this answer

answered Mar 27 at 16:47

st.mathst.math

1,268115

$endgroup$

add a comment | 

$begingroup$

To calculate $0^n-1$, you divided by $0$, which is undefined. Notice that $0^n=0cdots0=0$ (with $n$ factors); there is no division involved here.

share|cite|improve this answer

answered Mar 27 at 16:47

st.mathst.math

1,268115

$endgroup$

share|cite|improve this answer

answered Mar 27 at 16:47

st.mathst.math

1,268115

answered Mar 27 at 16:47

st.mathst.math

1,268115

answered Mar 27 at 16:47

st.mathst.math

1,268115