35. Show that 5 + 3 is an irrational number, given that positive square root of 15 is an
irrational number.
Share
35. Show that 5 + 3 is an irrational number, given that positive square root of 15 is an
irrational number.
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Step-by-step explanation:
This proof uses the unique prime factorisation theorem that every positive integer has a unique factorisation as a product of positive prime numbers.
Suppose √15=pq for some p,q∈N . and that andq are the smallest such positive integers.
Then p2=15q2 The right hand side has factors of 3 and 5 , so p2 must be divisible by 3 and by 5 . By the unique prime factorisationtheorem, p must also be divisible by 3 and 5 .So p=3⋅5.k=15k for some k∈N .
Then we have:15q2=p2=
(15k)2=15⋅(15k2)
Divide both ends by 15
to find:q
2=15k2
So 15=q2k2 and √15=qk
Now
k<q<p
contradicting our assertion that
p,q
is the smallest pair of values such that
√15=pq .
So our initial assertion was false and there is no such pair of integers.
Step-by-step explanation:
This proof uses the unique prime factorisation theorem that every positive integer has a unique factorisation as a product of positive prime numbers.
Suppose √15=pq for some p,q∈N . and that andq are the smallest such positive integers.
Then p2=15q2 The right hand side has factors of 3 and 5 , so p2 must be divisible by 3 and by 5 . By the unique prime factorisationtheorem, p must also be divisible by 3 and 5 .So p=3⋅5.k=15k for some k∈N .
Then we have:15q2=p2=
(15k)2=15⋅(15k2)
Divide both ends by 15
to find:q
2=15k2
So 15=q2k2 and √15=qk
Now
k<q<p
contradicting our assertion that
p,q
is the smallest pair of values such that
√15=pq .
So our initial assertion was false and there is no such pair of integers.
Thanks!!!