Find a representation as a product of \(a^{2n+1} + b^{2n+1}\) for general \(a,b,n\).
Find a representation as a product of \(a^n - b^n\) for general \(a,b,n\).
Let \(a,b,c,d\) be positive real numbers. Prove that \((a+b)\times(c+d) = ac+ad+bc+bd\). Find both algebraic solution and geometric interpretation.
Let \(a,b,c,d\) be positive real numbers such that \(a\geq b\) and \(c\geq d\). Prove that \((a-b)\times(c-d) = ac-ad-bc+bd\). Find both algebraic solution and geometric interpretation.
Using the area of a rectangle prove that \(a\times b=b\times a\).
Which of the following numbers are divisible by \(11\) and which are not? \[121,\, 143,\, 286, 235, \, 473,\, 798, \, 693,\, 576, \,748\] Can you write down and prove a divisibility rule which helps to determine if a three digit number is divisible by \(11\)?
Do there exist two numbers such that their sum, quotient and product would be all equal to each other?
Recall that \((n+1)^2=n^2+2n+1\). Subtracting \(2n+1\) from both sides gives \((n+1)^2-(2n+1)=n^2\). Now subtract \(n(2n+1)\) from both sides to obtain \((n+1)^2-(2n+1)-n(2n+1)=n^2-n(2n+1)\). Notice that the left-hand side can be rewritten as \((n+1)^2-(n+1)(2n+1)\), so we have \((n+1)^2-(n+1)(2n+1)=n^2-n(2n+1)\).
Next, add \(\frac{(2n+1)^2}{4}\) to both sides. This allows us to complete the square on each side, giving \(((n+1)-\frac{2n+1}{2})^2=(n-\frac{2n+1}{2})^2\).
Taking square roots of both sides leads to \((n+1)-\frac{2n+1}{2}=n-\frac{2n+1}{2}\). Adding \(\frac{2n+1}{2}\) to both sides produces \(n+1=n\), which simplifies to \(1=0\).
Let’s prove that \(1\) is the smallest positive real number: Assume the contrary and let \(x\) be the smallest positive real number. If \(x>1\) then \(1\) is smaller, thus \(x\) is not the smallest. If \(x<1,\) then \(\frac{x}{2}<x\) so \(x\) can not be the smallest either. Then \(x\) can only be equal to \(1\).
Find the last two digits of the number \[33333333333333333347^4 - 11111111111111111147^4\]