Use proof by contradiction to show that there is no value of n for which 6n ends with the digit zero.

Consider the numbers of the form 4^n where n is a natural number. Check whether there is any value of n for which 4^n ends with zero?

Use proof by contradiction to prove that if for an integer a, a^2 is even, then so is a.

Show that there is no positive integer, n for which sqrt(n-1) + sqrt(n+1) is rational .

Use proof by contradiction to prove that if for an integer a, a^2 is divisible by 3, then a is divisible by 3.

Let r be a rational number and x be an irrational number. Use proof by contradiction to show that r+x is an irrational number.

Can the number 6^n , n being a natural number, end with the digit 5? Give reason.

An n - digit number is a positive number with exactly n digits. Nine hundred distinct n - digit numbers are to be formed using only the three digits 2,5 and 7 . The smallest value of n for which this is possible is :

For any 'n', 6^(n) - 5^(n) ends with

Check the validity of the statements given below by the method given against it. (i) p: The sum of an irrational number and a rational number is irrational (by contradiction method). (ii) q: If n is a real number with n gt 3 , then n^(2)gt9 (by contradiction method).

Suppose a+b=c+d, and alt c . Use proof by contradiction to show b gt d .