proof by contradiction part 2

Question 1

Prove bt contradiction that there is npc smallest6 positive rational number

Answer 1

Assume there is a smallest positive rational number r let x = r divided but 10 
As r is a postive rational number x is too but x is smaller than r. This contracts the statement that r is the smallest positive rational number

Question 2

Prove by contradiction that there are infinitely many prime numbers

Answer 2

Assume there are a finite number of prime numbers
Let these prime numbers be in increasomh order
Now consider n = them times together
Dividingn by any of the prime numbers listed gives a reminder of 1
Hence none of the prime numbers are factors of n
This means that n must either be a prime number or have a prime factor which is not in the finite list of primes given along
This contradicts therefore

Question 3

Prove byt contrdiction that the hypotenuse of a right angled triangle is smaller in length than the sum of the other two sides:

Answer 3

Assume that the hypotenuse of a right angled triangle is greater than or equal to the sum of the other two sides
All a,b and c are positive
So c^2> = a +b squared (do that)
Now pythagorus says acquired plus b square = c squared but a and b are both positive so 2ab > 0 not less than or equal to 0
Therefore it must be tried that the hypotenuse of a right angled triangle us smaller in length than the sum of the other two sides

Question 4

Prove by contrition that sins + cost is qual tan or greater to 1 for 0 to 90

Answer 4

Assume that six + cost is less than 1 for 0 to 90
For 0 to 90 less than and eqiual tp sin x > 0 and cos x>0
Hence hence sin x + cos x > 0
This we have 0< six + cost <1 now sin + cos x square do it
Identitos of both of them squared = 1
We have 2sinxcos + 1 < 1
2sinxcosx<0
But as six >0 and cos x >0 this cannot be true

Question 5

Prove by contradiction that there exists no integers a and b for which 21a + 14b = 1

Answer 5

Assume there exists integers a and b for which 21a and 14b = 1
Divide by 7 as it is the highest common factor to get 3a + 2b =1 divided by 7
But as 3a and 2b integers 3a + 2b should be an integer but it is 1 divided by 7 so this must be a contridtcion.