Sure! Here’s the transcription for an educational website:---**Mathematical Proof Exercises**1. **Exercise 6:** Use a direct proof to show that the product of two odd numbers is odd.2. **Exercise 8:** Prove that if \( n \) is a perfect square, then \( n + 2 \) is not a perfect square.3. **Exercise 12:** Prove or disprove that the product of a nonzero rational number and an irrational number is irrational.---These exercises are designed to help develop skills in mathematical reasoning and proof techniques.

We know that every odd numbers are of the type 2n-1 , where n is an integer.6. Suppose 2n-1 and 2m-1 are two odd numbers where n,m are integers. Now (2n-1)(2m-1)=4mn-2n-2m+1⇒(2n-1)(2m-1)=2(2mn-n-m+1)-1⇒(2n-1)(2m-1)=2p-1 where p=2mn-n-m+1 is an integer Which shows that (2n-1)(2m-1) is an odd number. Hence the product of two odd numbers is odd.8. Suppose n is a perfect square. Then n=m2 where m≥0 integer. If possible suppose that n+2 is a perfect square. Then there exists natural a number r such that n+2=r2 Now r2-m2=2⇒(r+m)(r-m)=2⇒r+m and r-m are divisor of 2⇒ r=1 and m=0r=0 and m=1r=1 and m=1r=2 and m=0r=0 and m=2 these are all possible cases. We saw that for all the cases we did not get r2-m2=2, which is a contradiction. Hence if n is a perfect square, then n+2 is not perfect square. 12. Suppose r=mn; m≠0 be a non zero rational number and x is an irrational number. If possible suppose that the product of a non zero rational and an irrational number is a rational number. Then we get rx is a rational number pq say. Therefore rx=pq⇒x=pqr⇒x=pnqm⇒x=st where s=pn and t=qm≠0 are integers. as q≠0, m≠0 Which shows that x is a rational number. Which is a contradiction. Hence the product of a non zero rational and an irrational number is irrational.

Math

Advanced Math

6. Use a direct proof to show that the product of two odd numbers is odd. 8. Prove that if n is a perfect square, then n + 2 is not a perifect square 12. Prove or disprove that the product of a nonzero rational number and an irrational number is irrational.