The following require both an answer to the true false quest

The following require both an answer to the true false question and a justification (either a proof or a counter example) for why. (a)True or False and why: If a and b are both irrational then ab is irrational. (b)True or False and why: If a and b are both irrational then a 4- b is irrational. (c)True or False and why: If a and b are both irrational then a^b is irrational. Prove that there are infinitely many primes of the form 4n + 3. Prove that if a, b, c Z and a^2 + b^2 = c^2 then abc is even. Prove that if a, b Z then a^2 + 9b notequalto 3. Let a, b, p Z with p prime. Prove that if p|a and p|b then p a - b^2

Solution

a)

False

Let, a=b=sqrt{2}

ab=2 which is rational but a and b are irrational

b)

False

Let, a=sqrt{2},b=-sqrt{2} , so both a,b irrational

a+b=0 which is rational

c)

False

Consider this

e^{ln 2}=2

e is irrational, ln 2 is irrational. But 2 is rational

1.

Case 1. a and b of same parity. Then, a ,b both odd or a,b both even

Then, a^2+b^2 is sum of two integers of same partiy hence the sum is even ie c is even

Hence, abc is even

Case 2. a and b of opposite parity ie one ie even and other odd

Since, one of a and b is even . So, abc is even

2.

True

Assume true

So we have

a^2+9b=3

a^2=3-9b=3(1-3b)

So, a^2 is multiple of 3

But 3 is prime, so a is a multiple of 3

Hence, 3 | 1-3b which is a contradiction

Hence, so a,b exist so that

a^2+9b=3

3.

Assume, p|a-b^2

a-b^2=kp for some integer k

p|a so : a=mp for some integer m

Hence, mp-kp=b^2

So, b^2 is a multiple of p

p|b^2. But, p is prime, so p|b

which is a contradiction

Hence, p does nto divide a-b^2