Greetings Everyone

Can someone please assist. Found this question in unit one assessment but am not sure how to go about it.

The question is ,"what decimal number has the binary representation 1011010?.

Essentially, the question you need to answer is how you would write the binary number `1011010`

normally – that is, in our base-10 or “decimal” system.

Binary numbers are pretty easy *once you understand them*, but a little tricky to understand at first – at least, they were for me.

The main resource we have on this topic is the Khan Academy video in CS101 1.2.2. I’ll embed it below, too. But it could also be useful to seek out some other explanations, like this one from Math Is Fun.

Here is the explanation that makes the most sense to me (but I invite anyone to critique my method ):

Suppose we have an unknown five digit binary number. Starting from the right, each digit in the binary number is an increment of a power of 2.

2^{4} 2^{3} 2^{2} 2^{1} 2^{0}

Since this is binary, for each digit, you only get a choice of `0`

or `1`

. For each digit, then, you can have `zero`

amount of that particularly power of two, or you can have `one`

of that particular power of two.

Let’s suppose we have a number, then, like `1 1 1 1 1`

. That’s like saying that I have one of each kind of power of two. I have one of 2^{4}, one 2^{3}, one 2^{2}, one 2^{1}, and one 2^{0}.

Put another way, that means I have one 16, one 8, one 4, one 2, and one 1 (since 2^{0} equals 1).

Just add them all up:

`16 + 8 + 4 + 2 + 1 = 31`

So the binary number `1 1 1 1 1`

is equal to the decimal number `31`

.

Another quick example. Suppose I have the binary number `1 1 0 1 0`

. Now I have one 2^{4}, one 2^{3}, **no** 2^{2}, one 2^{1}, and **no** 2^{0}.

I can add these up, too:

`16 + 8 + [0] + 2 + [0] = 26`

Okay, last one: `1 0 1 0 1`

. That will just be this:

`16 + [0] + 4 + [0] + 1 = 21`

I’ve only dealt above with 5 digit binary numbers. The problem above is asking you to look at a 7 digit binary number. That means we need to tack on two additional powers of 2, like so:

2^{6} 2^{5} 2^{4} 2^{3} 2^{2} 2^{1} 2^{0}

All you really need to do is to convert *each digit* of the binary number into an equivalent base-10 (decimal) number, and then add up all the numbers you get.

Decimal numbers really work the same way, except we have ten digits (0 through 9) instead of two (0, 1) and we deal with powers of 10 instead of powers of 2. The **Math Is Fun** link above does a good job showing that, I think.

Here’s that Khan video I promised, the same one we use in section 1.2.2:

@Sean

Thank you very much for the time and trouble you took in giving the answer to my question. The information you gave was very helpful and indeed enriching. I loved the Math is Fun. It is quite a good resource on the subject. I greatly appreciate.

Hi, this is an easy way that I learned binary conversions and this is to supplement the video in the course.

These are examples from a you tube video that I watched called “How to Convert Binary to Decimal Tutorial (The Easy Way)”

Binary to decimal:

Example: 10000100 to 132

You put the binary numbers in the above example under each decimal number

128 64 32 16 8 4 2 1

1 0 0 0 0 1 0 0

So, after you have placed the binary numbers under the decimal numbers, then you will add the decimal numbers corresponding to the ones only not the zeros.

Ex. There is a 1 corresponding to 128, there is a 1 corresponding to the number 4 and. So you will get the total which is your decimal number.

128 +4= 132

132 will be the decimal number

Decimal to Binary

Example: 27 to 11011

In this example you turn decimal into binary using the decimal number 27 which is 11011 binary.

You will put the binary numbers under the corresponding decimal number

128 64 32 16 8 4 2 1

1 1 0 1 1

Ex. You will add the 16+8+2+1 you only add the decimal numbers together with a one

Under them not the zeros.

16 +8+2+1= 27.

So, 27 is your decimal number.

This is some very basic steps to help get you introduced to the binary decimal conversion.

I found this technique to be the easiest so far.

Hope this helps.