About question of fundamental theorem of arithmetic
A product of nonzero integers whose absolute values are $<p$ will have the property that all its prime factors are $≤p−1$.
I know that Composite Number has Prime Factor not Greater Than its Square Root. But how this solve the above problem.
number-theory
asked Oct 30 '18 at 14:07
ninja hatori
5721421
add a comment |
A product of nonzero integers whose absolute values are $<p$ will have the property that all its prime factors are $≤p−1$.
I know that Composite Number has Prime Factor not Greater Than its Square Root. But how this solve the above problem.
number-theory
asked Oct 30 '18 at 14:07
ninja hatori
5721421
1
Use contradiction.
– Math_QED
Oct 30 '18 at 14:08
How? can you explain it? @Math_QED
– ninja hatori
Oct 30 '18 at 14:13
add a comment |
A product of nonzero integers whose absolute values are $<p$ will have the property that all its prime factors are $≤p−1$.
I know that Composite Number has Prime Factor not Greater Than its Square Root. But how this solve the above problem.
number-theory
asked Oct 30 '18 at 14:07
ninja hatori
5721421
A product of nonzero integers whose absolute values are $<p$ will have the property that all its prime factors are $≤p−1$.
I know that Composite Number has Prime Factor not Greater Than its Square Root. But how this solve the above problem.
number-theory
number-theory
asked Oct 30 '18 at 14:07
ninja hatori
5721421
asked Oct 30 '18 at 14:07
ninja hatori
5721421
asked Oct 30 '18 at 14:07
ninja hatori
5721421
asked Oct 30 '18 at 14:07
ninja hatori
5721421
asked Oct 30 '18 at 14:07
ninja hatori
5721421
5721421
1
Use contradiction.
– Math_QED
Oct 30 '18 at 14:08
How? can you explain it? @Math_QED
– ninja hatori
Oct 30 '18 at 14:13
add a comment |
1
Use contradiction.
– Math_QED
Oct 30 '18 at 14:08
How? can you explain it? @Math_QED
– ninja hatori
Oct 30 '18 at 14:13
1
1
Use contradiction.
– Math_QED
Oct 30 '18 at 14:08
Use contradiction.
– Math_QED
Oct 30 '18 at 14:08
How? can you explain it? @Math_QED
– ninja hatori
Oct 30 '18 at 14:13
How? can you explain it? @Math_QED
– ninja hatori
Oct 30 '18 at 14:13
add a comment |
1 Answer
1
active
oldest
votes
It suffices to prove this for nonnegative numbers.
Note that $a, b < p implies a, b leq p-1$. Try to look at an equal factorization in power of primes (nonnegative exponents), and let $p_k$ be the greatest prime factor of either $a$ or $b$. Then by the Fundamental Theorem of Arithmetic, their products have the same greatest prime factor, which either satisfies $p_k = p-1$ or $p_k leq sqrt{p-1} leq p-1$. Hence all the prime factors $p_i$ are less or equal to $p-1$.
answered Oct 30 '18 at 14:48
Lucas Henrique
985314
add a comment |
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1 Answer
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It suffices to prove this for nonnegative numbers.
Note that $a, b < p implies a, b leq p-1$. Try to look at an equal factorization in power of primes (nonnegative exponents), and let $p_k$ be the greatest prime factor of either $a$ or $b$. Then by the Fundamental Theorem of Arithmetic, their products have the same greatest prime factor, which either satisfies $p_k = p-1$ or $p_k leq sqrt{p-1} leq p-1$. Hence all the prime factors $p_i$ are less or equal to $p-1$.
answered Oct 30 '18 at 14:48
Lucas Henrique
985314
add a comment |
It suffices to prove this for nonnegative numbers.
Note that $a, b < p implies a, b leq p-1$. Try to look at an equal factorization in power of primes (nonnegative exponents), and let $p_k$ be the greatest prime factor of either $a$ or $b$. Then by the Fundamental Theorem of Arithmetic, their products have the same greatest prime factor, which either satisfies $p_k = p-1$ or $p_k leq sqrt{p-1} leq p-1$. Hence all the prime factors $p_i$ are less or equal to $p-1$.
answered Oct 30 '18 at 14:48
Lucas Henrique
985314
add a comment |
It suffices to prove this for nonnegative numbers.
Note that $a, b < p implies a, b leq p-1$. Try to look at an equal factorization in power of primes (nonnegative exponents), and let $p_k$ be the greatest prime factor of either $a$ or $b$. Then by the Fundamental Theorem of Arithmetic, their products have the same greatest prime factor, which either satisfies $p_k = p-1$ or $p_k leq sqrt{p-1} leq p-1$. Hence all the prime factors $p_i$ are less or equal to $p-1$.
answered Oct 30 '18 at 14:48
Lucas Henrique
985314
It suffices to prove this for nonnegative numbers.
Note that $a, b < p implies a, b leq p-1$. Try to look at an equal factorization in power of primes (nonnegative exponents), and let $p_k$ be the greatest prime factor of either $a$ or $b$. Then by the Fundamental Theorem of Arithmetic, their products have the same greatest prime factor, which either satisfies $p_k = p-1$ or $p_k leq sqrt{p-1} leq p-1$. Hence all the prime factors $p_i$ are less or equal to $p-1$.
answered Oct 30 '18 at 14:48
Lucas Henrique
985314
edited Nov 27 '18 at 15:04
answered Oct 30 '18 at 14:48
Lucas Henrique
985314
answered Oct 30 '18 at 14:48
Lucas Henrique
985314
answered Oct 30 '18 at 14:48
Lucas Henrique
985314
985314
add a comment |
add a comment |
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Use contradiction.
– Math_QED
Oct 30 '18 at 14:08
How? can you explain it? @Math_QED
– ninja hatori
Oct 30 '18 at 14:13