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2=1? Is that true?

Often we see the evidence mathematically that 2=1. They show very logically. Some people who cannot know this mistake then they blindly assume that math is flawed.

I believe that mathematics is one of the perfect sciences. Because until now I have not found a defect in this science. Which I love, this science is connected to each other. Some people even consider that mathematics is the language of Deity.

Back to the topic of conversation. Ever seen evidence like this?

    \begin{eqnarray*} a & = & b\\ a^{2} & = & ab\\ a^{2}-b^{2} & = & ab-b^{2}\\ \left(a+b\right)\left(a-b\right) & = & b\left(a-b\right)\\ a+b & = & b\\ 2b & = & b\\ 2 & = & 1\end{eqnarray*}

Or something like this?

    \begin{eqnarray*} a & = & b\\ a^{2} & = & ab\\ a^{2}+a^{2} & = & a^{2}+ab\\ 2a^{2} & = & a^{2}+ab\\ 2a^{2}-2ab & = & a^{2}+ab-2ab\\ 2a^{2}-2ab & = & a^{2}-ab\\ 2\left(a^{2}-ab\right) & = & \left(a^{2}-ab\right)\\ 2 & = & 1\end{eqnarray*}

Lots of evidence like this is scattered on the internet. If we do not look closely, we will be caught in logical traps and as if they are true.

So, where is the error of evidence above?

It’s simple. The error of the above evidence is in the step that makes \frac{0}{0}=1. And we know that \frac{0}{0} is undefined.

I do not see the \frac{0}{0} . Where is it? (more…)

Proof / Disclaimer With Example

I think my post is a bit long and a lot of hassle lately. Today I will make a basic article. This is for website refreshment as well. Happy reading.

Most people use the equivalent method to prove mathematical statements. Even to prove the error of a statement. Though there are other ways to prove error statement, ie with an example of denial. This is considered sufficient because a statement must be perfect. Also because the truth must be perfect in mathematics.

Suppose there is a statement “If n\in\mathbb{N} then n^{2}+n+1 is prime”. Suppose we want to deny the statement. We have a presumption that n^{2}+n+1 not always prime. Then simply indicated there is a number n=k such that k^{2}+k+1 is not prime. For this case, we take n=4 then n^{2}+n+1=4^{2}+4+1=21 is not prime. This example is sufficient to prove that the above statement is false.

Thanks for reading.

How To Proof Set A Is Equal To Set B

Some people are wrong ways to prove that set A is equal to set B. I also experienced it when I started learning math. Though the concept of set A is equal to set B very simple. We only have to prove that set A is a subset of set B and set B is a subset of set A. Of course the basic is very easy to understand.

Then, how to prove the set A is a subset of the set B, vice versa?

The way to prove set A is a subset of set B is by taking any element on set A and then prove that element is a element of set B. It can be easily understood that the taking of any element can be representative of all element of the set. Mathematically like this :

    \[A=B\Longleftrightarrow\left(A\subseteq B\right)\wedge\left(B\subseteq A\right)\]


    \[A\subseteq B\Longleftrightarrow\left(\left(x\in A\right)\Rightarrow\left(x\in B\right)\right)\]

Example : (more…)

An Example of Using Mathematical Induction

In this post will be given some examples as well as some banks about the problem of mathematical induction. You can learn from some examples and can then practice from some unanswered questions.

Example 1.

Prove that for all n\in\mathbb{N} then :


Answer :

Step 1. (proving the statement is true for n=1)

For n=1

Then 1=\frac{1}{2}1\left(1+1\right) \longrightarrow (is true)

Step 2. (assume correct for n=k)

Assume correct for any n=k then it is true that


Step 3.

(Prove true for n=k+1 with capital assumptions in step 2. So we have to prove that 1+2+3+\cdots+k+\left(k+1\right)=\frac{1}{2}\left(k+1\right)\left(k+2\right))

For n=k+1



    =\frac{1}{2}k\left(k+1\right)+\left(k+1\right)<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="http://www.infinite-set.com/wp-content/ql-cache/quicklatex.com-024ed878b853acf437a4cf5cc647bdb3_l3.png" height="75" width="582" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[\longrightarrow$ we have capital assumptions that $1+2+3+\ldots+k=\frac{1}{2}k\left(k+1\right)$ $=\frac{k\left(k+1\right)+2\left(k+1\right)}{2}$ $=\frac{k^{2}+k+2k+2}{2}$ $=\frac{k^{2}+3k+2}{2}$ $=\frac{\left(k+1\right)\left(k+2\right)}{2}$ $=\frac{1}{2}\left(k+1\right)\left(k+2\right)\]" title="Rendered by QuickLaTeX.com"/>\blacksquare

I hope the explanation in the previous article and with the example above that I write a description in detail you can understand easily. So in the next example we will work with steps without much lengthy explanation because the basics are the same.

Example 2. (more…)