$a = d$ implies $a^b = d^b$
$begingroup$
Prove that $a = d$ implies $a^b = d^b$, where $a, d$ are arbitrary
nonnegative integers and $b$ is any positive integer.
If I could use division I think it could be something like that:
$a^b / d^b = a ^{b-b} = a^0 = 1$ (assuming $a = d$), but I'm trying to figure out how to prove this using only multiplication and addition properties (natural numbers).
Here's my idea:
If a = b then there exists an integer $k$ such that $x = ak$ and $x = bk$, that also implies $(a^b)k = (d^b) k$ and $a^b = d^b$ by using the cancellation law of multiplication.
proof-verification natural-numbers
asked Dec 7 '18 at 17:35
adriana634adriana634
436
$endgroup$
add a comment |
$begingroup$
Prove that $a = d$ implies $a^b = d^b$, where $a, d$ are arbitrary
nonnegative integers and $b$ is any positive integer.
If I could use division I think it could be something like that:
$a^b / d^b = a ^{b-b} = a^0 = 1$ (assuming $a = d$), but I'm trying to figure out how to prove this using only multiplication and addition properties (natural numbers).
Here's my idea:
If a = b then there exists an integer $k$ such that $x = ak$ and $x = bk$, that also implies $(a^b)k = (d^b) k$ and $a^b = d^b$ by using the cancellation law of multiplication.
proof-verification natural-numbers
asked Dec 7 '18 at 17:35
adriana634adriana634
436
$endgroup$
add a comment |
$begingroup$
Prove that $a = d$ implies $a^b = d^b$, where $a, d$ are arbitrary
nonnegative integers and $b$ is any positive integer.
If I could use division I think it could be something like that:
$a^b / d^b = a ^{b-b} = a^0 = 1$ (assuming $a = d$), but I'm trying to figure out how to prove this using only multiplication and addition properties (natural numbers).
Here's my idea:
If a = b then there exists an integer $k$ such that $x = ak$ and $x = bk$, that also implies $(a^b)k = (d^b) k$ and $a^b = d^b$ by using the cancellation law of multiplication.
proof-verification natural-numbers
asked Dec 7 '18 at 17:35
adriana634adriana634
436
$endgroup$
Prove that $a = d$ implies $a^b = d^b$, where $a, d$ are arbitrary
nonnegative integers and $b$ is any positive integer.
If I could use division I think it could be something like that:
$a^b / d^b = a ^{b-b} = a^0 = 1$ (assuming $a = d$), but I'm trying to figure out how to prove this using only multiplication and addition properties (natural numbers).
Here's my idea:
If a = b then there exists an integer $k$ such that $x = ak$ and $x = bk$, that also implies $(a^b)k = (d^b) k$ and $a^b = d^b$ by using the cancellation law of multiplication.
proof-verification natural-numbers
proof-verification natural-numbers
asked Dec 7 '18 at 17:35
adriana634adriana634
436
asked Dec 7 '18 at 17:35
adriana634adriana634
436
asked Dec 7 '18 at 17:35
adriana634adriana634
436
asked Dec 7 '18 at 17:35
adriana634adriana634
436
asked Dec 7 '18 at 17:35
adriana634adriana634
436
436
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You're overthinking it. We proceed by induction on $b$. The case $b=1$ is trivial, so assume it works when $b=k$. Then $a^{k+1}=a^ka=d^kd=d^{k+1}$ completes the inductive step.
answered Dec 7 '18 at 17:39
J.G.J.G.
24.1k22539
$endgroup$
add a comment |
$begingroup$
There must be something I don't understand. Are you sure you have asked the question you intended?
The equals sign in the assumption $a=d$ means that "$a$" and "$d$" are essentially just different names for the same number. So you can substitute one for the other in any formula. There is nothing to prove.
answered Dec 7 '18 at 18:50
Ethan BolkerEthan Bolker
42.1k548111
$endgroup$
$begingroup$
You took the words right out of my mouth.
$endgroup$
– Shubham Johri
Dec 7 '18 at 21:00
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You're overthinking it. We proceed by induction on $b$. The case $b=1$ is trivial, so assume it works when $b=k$. Then $a^{k+1}=a^ka=d^kd=d^{k+1}$ completes the inductive step.
answered Dec 7 '18 at 17:39
J.G.J.G.
24.1k22539
$endgroup$
add a comment |
$begingroup$
You're overthinking it. We proceed by induction on $b$. The case $b=1$ is trivial, so assume it works when $b=k$. Then $a^{k+1}=a^ka=d^kd=d^{k+1}$ completes the inductive step.
answered Dec 7 '18 at 17:39
J.G.J.G.
24.1k22539
$endgroup$
add a comment |
$begingroup$
You're overthinking it. We proceed by induction on $b$. The case $b=1$ is trivial, so assume it works when $b=k$. Then $a^{k+1}=a^ka=d^kd=d^{k+1}$ completes the inductive step.
answered Dec 7 '18 at 17:39
J.G.J.G.
24.1k22539
$endgroup$
You're overthinking it. We proceed by induction on $b$. The case $b=1$ is trivial, so assume it works when $b=k$. Then $a^{k+1}=a^ka=d^kd=d^{k+1}$ completes the inductive step.
answered Dec 7 '18 at 17:39
J.G.J.G.
24.1k22539
answered Dec 7 '18 at 17:39
J.G.J.G.
24.1k22539
answered Dec 7 '18 at 17:39
J.G.J.G.
24.1k22539
answered Dec 7 '18 at 17:39
J.G.J.G.
24.1k22539
24.1k22539
add a comment |
add a comment |
$begingroup$
There must be something I don't understand. Are you sure you have asked the question you intended?
The equals sign in the assumption $a=d$ means that "$a$" and "$d$" are essentially just different names for the same number. So you can substitute one for the other in any formula. There is nothing to prove.
answered Dec 7 '18 at 18:50
Ethan BolkerEthan Bolker
42.1k548111
$endgroup$
$begingroup$
You took the words right out of my mouth.
$endgroup$
– Shubham Johri
Dec 7 '18 at 21:00
add a comment |
$begingroup$
There must be something I don't understand. Are you sure you have asked the question you intended?
The equals sign in the assumption $a=d$ means that "$a$" and "$d$" are essentially just different names for the same number. So you can substitute one for the other in any formula. There is nothing to prove.
answered Dec 7 '18 at 18:50
Ethan BolkerEthan Bolker
42.1k548111
$endgroup$
$begingroup$
You took the words right out of my mouth.
$endgroup$
– Shubham Johri
Dec 7 '18 at 21:00
add a comment |
$begingroup$
There must be something I don't understand. Are you sure you have asked the question you intended?
The equals sign in the assumption $a=d$ means that "$a$" and "$d$" are essentially just different names for the same number. So you can substitute one for the other in any formula. There is nothing to prove.
answered Dec 7 '18 at 18:50
Ethan BolkerEthan Bolker
42.1k548111
$endgroup$
There must be something I don't understand. Are you sure you have asked the question you intended?
The equals sign in the assumption $a=d$ means that "$a$" and "$d$" are essentially just different names for the same number. So you can substitute one for the other in any formula. There is nothing to prove.
answered Dec 7 '18 at 18:50
Ethan BolkerEthan Bolker
42.1k548111
answered Dec 7 '18 at 18:50
Ethan BolkerEthan Bolker
42.1k548111
answered Dec 7 '18 at 18:50
Ethan BolkerEthan Bolker
42.1k548111
answered Dec 7 '18 at 18:50
Ethan BolkerEthan Bolker
42.1k548111
42.1k548111
$begingroup$
You took the words right out of my mouth.
$endgroup$
– Shubham Johri
Dec 7 '18 at 21:00
add a comment |
$begingroup$
You took the words right out of my mouth.
$endgroup$
– Shubham Johri
Dec 7 '18 at 21:00
$begingroup$
You took the words right out of my mouth.
$endgroup$
– Shubham Johri
Dec 7 '18 at 21:00
$begingroup$
You took the words right out of my mouth.
$endgroup$
– Shubham Johri
Dec 7 '18 at 21:00
add a comment |
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