Question

35. Show that 5 + 3 is an irrational number, given that positive square root of 15 is an
irrational number.​

Answer 1

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.

Answer 2

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!!!