C Programming - Floating Point Issues - Discussion

5.375
here know 5=101
now for 0.375

0.375
*2
____________=0 .collect top to bottom
.
0.750 .
*2 .
____________=1 .
1.500

0.500
*2
____________=1
1.000

Nice answer vinoth thank you.

5.375

5 --> 101

0.375 * 2 = 0.75
0.75 * 2 = 1.5
0.5 * 2 = 1.0
Here the binary position like
.. 32 16 8 4 2 1
........................
1 1 = who is 1 add it that is 4+1=5. ok then remaing point matter

.375
in binary point conversion of divide by 2
and then LSB and MSB
0.375 *2 = 0.75 consider that 0
0.75 * 2 = 1.5 consider that 1
0.5 * 2 = 1.0 consider that 1
top to down
that is 011
and 5 is
101.011

Q.The binary equalent of 5.375 is ..........
ans1)
5 binary equalent is 101
0.375 binary equalent is 3-011
7-111
5-101
take the first bit of 3,7,5 that is 011
so total correct ans is 101.011

The above ans like this also..
2/375(1)
2/187(1)
3/93(0)
2/31(1)
3/15(0)
2/5(1)
2/2(0)
1
finally result is:10101111->101.011

@Susmita

Nice explanation.

Thank you surya.

@surya.

Your explanation is nice one.

In decimal numbers, the digits have values that are powers of ten. Starting at the implied decimal point and working to the left, the number 403 is equal to:
3 x 100 + 0 x 101 + 4 x 102 = 3x1 + 0x10 + 4x100 = 3 + 0 + 400

In binary, the digits have values that are powers of two. Working to the left as before, the number 101 is equal to:
1 x 20 + 0 x 21 + 1 x 22 = 1x1 + 0x2+ 1x4 = (in decimal) 1 + 0 + 4 = 5
So, 1111 1111 = 1x1 + 1x2 + 1x4 + 1x8 + 1x16 + 1x32 + 1x64 + 1x128 = 255.

This also works to the right of the decimal point, but the powers of two are negative (i.e., fractions), so 101.011 in binary is:
1x1/8 + 1x1/4 + 0x1/2 + 1x1 + 0x2 + 1x4 = 5.375

Where the first digit to the right of the decimal point is 2-1 (one-half, or 1/2), the next is 2-2 (1/4), then 2-3 (1/8), etc.

Why should we write only the decimal part from botom to top, it seems some more explanation is need to be there.