Using mathematical induction show that \(2^n>n\) for all natural numbers \(n\).
Illustrate with a picture
(a) \((a-b)^2 = a^2 - 2ab + b^2\),
(b) \(a^2 - b^2 = (a-b)(a+b)\),
(c) \((a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc\).
Suppose \(a>b\). Explain using the number line why
(a) \(a-c>b-c\), (b) \(2a>2b\).
Suppose \(a>b\) and \(c>d\). Prove that \(a+c>b+d\).
Using mathematical induction prove that \(2^n \geq n + 1\) for all natural numbers.
Circles and lines are drawn on the plane. They divide the plane into non-intersecting regions, see the picture below.
Show that it is possible to colour the regions with two colours in such a way that no two regions sharing some length of border are the same colour.
Consider a number consisting of \(3^n\) digits, all ones, such as 111 or 111111111 for example. Show that such a number with \(3^n\) digits is divisible by \(3^n\).
Numbers \(1,2,\dots,n\) are written on a whiteboard. In one go Louise is allowed to wipe out any two numbers \(a\) and \(b\), and write their sum \(a+b\) instead. Louise enjoys erasing the numbers, and continues the procedure until only one number is left on the whiteboard. What number is it? What if instead of \(a+b\) she writes \(a+b-1\)?
Prove that
(a) \[1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{1}{6} n (n+1)(2n+1)\]
(b) \[1^2 + 3^2 + 5^2 + \dots + (2n-1)^2 = \frac{1}{3} n (2n-1)(2n+1)\].
Is “I see what I eat” the same thing as “I eat what I see”?
To make it not so confusing let’s change the wording to make it more “mathematical”
“I see what I eat”=“If I eat it then I see it”
“I eat what I see”= “If I see it then I eat it”