Convert the Number 1.010 101 010 101 016 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number. Detailed Explanations
Number 1.010 101 010 101 016(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)
The first steps we'll go through to make the conversion:
Convert to binary (to base 2) the integer part of the number.
Convert to binary the fractional part of the number.
1. First, convert to binary (in base 2) the integer part: 1.
Divide the number repeatedly by 2.
Keep track of each remainder.
We stop when we get a quotient that is equal to zero.
- division = quotient + remainder;
- 1 ÷ 2 = 0 + 1;
2. Construct the base 2 representation of the integer part of the number.
Take all the remainders starting from the bottom of the list constructed above.
1(10) =
1(2)
3. Convert to binary (base 2) the fractional part: 0.010 101 010 101 016.
Multiply it repeatedly by 2.
Keep track of each integer part of the results.
Stop when we get a fractional part that is equal to zero.
- #) multiplying = integer + fractional part;
- 1) 0.010 101 010 101 016 × 2 = 0 + 0.020 202 020 202 032;
- 2) 0.020 202 020 202 032 × 2 = 0 + 0.040 404 040 404 064;
- 3) 0.040 404 040 404 064 × 2 = 0 + 0.080 808 080 808 128;
- 4) 0.080 808 080 808 128 × 2 = 0 + 0.161 616 161 616 256;
- 5) 0.161 616 161 616 256 × 2 = 0 + 0.323 232 323 232 512;
- 6) 0.323 232 323 232 512 × 2 = 0 + 0.646 464 646 465 024;
- 7) 0.646 464 646 465 024 × 2 = 1 + 0.292 929 292 930 048;
- 8) 0.292 929 292 930 048 × 2 = 0 + 0.585 858 585 860 096;
- 9) 0.585 858 585 860 096 × 2 = 1 + 0.171 717 171 720 192;
- 10) 0.171 717 171 720 192 × 2 = 0 + 0.343 434 343 440 384;
- 11) 0.343 434 343 440 384 × 2 = 0 + 0.686 868 686 880 768;
- 12) 0.686 868 686 880 768 × 2 = 1 + 0.373 737 373 761 536;
- 13) 0.373 737 373 761 536 × 2 = 0 + 0.747 474 747 523 072;
- 14) 0.747 474 747 523 072 × 2 = 1 + 0.494 949 495 046 144;
- 15) 0.494 949 495 046 144 × 2 = 0 + 0.989 898 990 092 288;
- 16) 0.989 898 990 092 288 × 2 = 1 + 0.979 797 980 184 576;
- 17) 0.979 797 980 184 576 × 2 = 1 + 0.959 595 960 369 152;
- 18) 0.959 595 960 369 152 × 2 = 1 + 0.919 191 920 738 304;
- 19) 0.919 191 920 738 304 × 2 = 1 + 0.838 383 841 476 608;
- 20) 0.838 383 841 476 608 × 2 = 1 + 0.676 767 682 953 216;
- 21) 0.676 767 682 953 216 × 2 = 1 + 0.353 535 365 906 432;
- 22) 0.353 535 365 906 432 × 2 = 0 + 0.707 070 731 812 864;
- 23) 0.707 070 731 812 864 × 2 = 1 + 0.414 141 463 625 728;
- 24) 0.414 141 463 625 728 × 2 = 0 + 0.828 282 927 251 456;
- 25) 0.828 282 927 251 456 × 2 = 1 + 0.656 565 854 502 912;
- 26) 0.656 565 854 502 912 × 2 = 1 + 0.313 131 709 005 824;
- 27) 0.313 131 709 005 824 × 2 = 0 + 0.626 263 418 011 648;
- 28) 0.626 263 418 011 648 × 2 = 1 + 0.252 526 836 023 296;
- 29) 0.252 526 836 023 296 × 2 = 0 + 0.505 053 672 046 592;
- 30) 0.505 053 672 046 592 × 2 = 1 + 0.010 107 344 093 184;
- 31) 0.010 107 344 093 184 × 2 = 0 + 0.020 214 688 186 368;
- 32) 0.020 214 688 186 368 × 2 = 0 + 0.040 429 376 372 736;
- 33) 0.040 429 376 372 736 × 2 = 0 + 0.080 858 752 745 472;
- 34) 0.080 858 752 745 472 × 2 = 0 + 0.161 717 505 490 944;
- 35) 0.161 717 505 490 944 × 2 = 0 + 0.323 435 010 981 888;
- 36) 0.323 435 010 981 888 × 2 = 0 + 0.646 870 021 963 776;
- 37) 0.646 870 021 963 776 × 2 = 1 + 0.293 740 043 927 552;
- 38) 0.293 740 043 927 552 × 2 = 0 + 0.587 480 087 855 104;
- 39) 0.587 480 087 855 104 × 2 = 1 + 0.174 960 175 710 208;
- 40) 0.174 960 175 710 208 × 2 = 0 + 0.349 920 351 420 416;
- 41) 0.349 920 351 420 416 × 2 = 0 + 0.699 840 702 840 832;
- 42) 0.699 840 702 840 832 × 2 = 1 + 0.399 681 405 681 664;
- 43) 0.399 681 405 681 664 × 2 = 0 + 0.799 362 811 363 328;
- 44) 0.799 362 811 363 328 × 2 = 1 + 0.598 725 622 726 656;
- 45) 0.598 725 622 726 656 × 2 = 1 + 0.197 451 245 453 312;
- 46) 0.197 451 245 453 312 × 2 = 0 + 0.394 902 490 906 624;
- 47) 0.394 902 490 906 624 × 2 = 0 + 0.789 804 981 813 248;
- 48) 0.789 804 981 813 248 × 2 = 1 + 0.579 609 963 626 496;
- 49) 0.579 609 963 626 496 × 2 = 1 + 0.159 219 927 252 992;
- 50) 0.159 219 927 252 992 × 2 = 0 + 0.318 439 854 505 984;
- 51) 0.318 439 854 505 984 × 2 = 0 + 0.636 879 709 011 968;
- 52) 0.636 879 709 011 968 × 2 = 1 + 0.273 759 418 023 936;
- 53) 0.273 759 418 023 936 × 2 = 0 + 0.547 518 836 047 872;
We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)
4. Construct the base 2 representation of the fractional part of the number.
Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:
0.010 101 010 101 016(10) =
0.0000 0010 1001 0101 1111 1010 1101 0100 0000 1010 0101 1001 1001 0(2)
5. Positive number before normalization:
1.010 101 010 101 016(10) =
1.0000 0010 1001 0101 1111 1010 1101 0100 0000 1010 0101 1001 1001 0(2)
The last steps we'll go through to make the conversion:
Normalize the binary representation of the number.
Adjust the exponent.
Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.
Normalize the mantissa.
6. Normalize the binary representation of the number.
Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:
1.010 101 010 101 016(10) =
1.0000 0010 1001 0101 1111 1010 1101 0100 0000 1010 0101 1001 1001 0(2) =
1.0000 0010 1001 0101 1111 1010 1101 0100 0000 1010 0101 1001 1001 0(2) × 20
7. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:
Sign 0 (a positive number)
Exponent (unadjusted): 0
Mantissa (not normalized):
1.0000 0010 1001 0101 1111 1010 1101 0100 0000 1010 0101 1001 1001 0
8. Adjust the exponent.
Use the 11 bit excess/bias notation:
Exponent (adjusted) =
Exponent (unadjusted) + 2(11-1) - 1 =
0 + 2(11-1) - 1 =
(0 + 1 023)(10) =
1 023(10)
9. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.
Use the same technique of repeatedly dividing by 2:
- division = quotient + remainder;
- 1 023 ÷ 2 = 511 + 1;
- 511 ÷ 2 = 255 + 1;
- 255 ÷ 2 = 127 + 1;
- 127 ÷ 2 = 63 + 1;
- 63 ÷ 2 = 31 + 1;
- 31 ÷ 2 = 15 + 1;
- 15 ÷ 2 = 7 + 1;
- 7 ÷ 2 = 3 + 1;
- 3 ÷ 2 = 1 + 1;
- 1 ÷ 2 = 0 + 1;
10. Construct the base 2 representation of the adjusted exponent.
Take all the remainders starting from the bottom of the list constructed above.
Exponent (adjusted) =
1023(10) =
011 1111 1111(2)
11. Normalize the mantissa.
a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.
b) Adjust its length to 52 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).
Mantissa (normalized) =
1. 0000 0010 1001 0101 1111 1010 1101 0100 0000 1010 0101 1001 1001 0 =
0000 0010 1001 0101 1111 1010 1101 0100 0000 1010 0101 1001 1001
12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:
Sign (1 bit) =
0 (a positive number)
Exponent (11 bits) =
011 1111 1111
Mantissa (52 bits) =
0000 0010 1001 0101 1111 1010 1101 0100 0000 1010 0101 1001 1001
The base ten decimal number 1.010 101 010 101 016 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 1111 - 0000 0010 1001 0101 1111 1010 1101 0100 0000 1010 0101 1001 1001
Sign (1 bit):
Exponent (11 bits):
0
621
611
601
591
581
571
561
551
541
531
52
Mantissa (52 bits):
0
510
500
490
480
470
461
450
441
430
420
411
400
391
380
371
361
351
341
331
321
310
301
290
281
271
260
251
240
231
220
210
200
190
180
170
161
150
141
130
120
111
100
91
81
70
60
51
41
30
20
11
0
Convert to 64 bit double precision IEEE 754 binary floating point representation standard
A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)
The latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation
How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard
Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:
- 1. If the number to be converted is negative, start with its the positive version.
- 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
- 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
- 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
- 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
- 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
- 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 - 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
- 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.
Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:
- 1. Start with the positive version of the number:
|-31.640 215| = 31.640 215
- 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
- division = quotient + remainder;
- 31 ÷ 2 = 15 + 1;
- 15 ÷ 2 = 7 + 1;
- 7 ÷ 2 = 3 + 1;
- 3 ÷ 2 = 1 + 1;
- 1 ÷ 2 = 0 + 1;
- We have encountered a quotient that is ZERO => FULL STOP
- 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:
31(10) = 1 1111(2)
- 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
- #) multiplying = integer + fractional part;
- 1) 0.640 215 × 2 = 1 + 0.280 43;
- 2) 0.280 43 × 2 = 0 + 0.560 86;
- 3) 0.560 86 × 2 = 1 + 0.121 72;
- 4) 0.121 72 × 2 = 0 + 0.243 44;
- 5) 0.243 44 × 2 = 0 + 0.486 88;
- 6) 0.486 88 × 2 = 0 + 0.973 76;
- 7) 0.973 76 × 2 = 1 + 0.947 52;
- 8) 0.947 52 × 2 = 1 + 0.895 04;
- 9) 0.895 04 × 2 = 1 + 0.790 08;
- 10) 0.790 08 × 2 = 1 + 0.580 16;
- 11) 0.580 16 × 2 = 1 + 0.160 32;
- 12) 0.160 32 × 2 = 0 + 0.320 64;
- 13) 0.320 64 × 2 = 0 + 0.641 28;
- 14) 0.641 28 × 2 = 1 + 0.282 56;
- 15) 0.282 56 × 2 = 0 + 0.565 12;
- 16) 0.565 12 × 2 = 1 + 0.130 24;
- 17) 0.130 24 × 2 = 0 + 0.260 48;
- 18) 0.260 48 × 2 = 0 + 0.520 96;
- 19) 0.520 96 × 2 = 1 + 0.041 92;
- 20) 0.041 92 × 2 = 0 + 0.083 84;
- 21) 0.083 84 × 2 = 0 + 0.167 68;
- 22) 0.167 68 × 2 = 0 + 0.335 36;
- 23) 0.335 36 × 2 = 0 + 0.670 72;
- 24) 0.670 72 × 2 = 1 + 0.341 44;
- 25) 0.341 44 × 2 = 0 + 0.682 88;
- 26) 0.682 88 × 2 = 1 + 0.365 76;
- 27) 0.365 76 × 2 = 0 + 0.731 52;
- 28) 0.731 52 × 2 = 1 + 0.463 04;
- 29) 0.463 04 × 2 = 0 + 0.926 08;
- 30) 0.926 08 × 2 = 1 + 0.852 16;
- 31) 0.852 16 × 2 = 1 + 0.704 32;
- 32) 0.704 32 × 2 = 1 + 0.408 64;
- 33) 0.408 64 × 2 = 0 + 0.817 28;
- 34) 0.817 28 × 2 = 1 + 0.634 56;
- 35) 0.634 56 × 2 = 1 + 0.269 12;
- 36) 0.269 12 × 2 = 0 + 0.538 24;
- 37) 0.538 24 × 2 = 1 + 0.076 48;
- 38) 0.076 48 × 2 = 0 + 0.152 96;
- 39) 0.152 96 × 2 = 0 + 0.305 92;
- 40) 0.305 92 × 2 = 0 + 0.611 84;
- 41) 0.611 84 × 2 = 1 + 0.223 68;
- 42) 0.223 68 × 2 = 0 + 0.447 36;
- 43) 0.447 36 × 2 = 0 + 0.894 72;
- 44) 0.894 72 × 2 = 1 + 0.789 44;
- 45) 0.789 44 × 2 = 1 + 0.578 88;
- 46) 0.578 88 × 2 = 1 + 0.157 76;
- 47) 0.157 76 × 2 = 0 + 0.315 52;
- 48) 0.315 52 × 2 = 0 + 0.631 04;
- 49) 0.631 04 × 2 = 1 + 0.262 08;
- 50) 0.262 08 × 2 = 0 + 0.524 16;
- 51) 0.524 16 × 2 = 1 + 0.048 32;
- 52) 0.048 32 × 2 = 0 + 0.096 64;
- 53) 0.096 64 × 2 = 0 + 0.193 28;
- We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
- 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:
0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)
- 6. Summarizing - the positive number before normalization:
31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)
- 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:
31.640 215(10) =
1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24
- 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:
Sign: 1 (a negative number)
Exponent (unadjusted): 4
Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0
- 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:
Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
100 0000 0011(2)
- 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):
Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0
Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100
- Conclusion:
Sign (1 bit) = 1 (a negative number)
Exponent (8 bits) = 100 0000 0011
Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100
Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100
Available Base Conversions Between Decimal and Binary Systems
Conversions Between Decimal System Numbers (Written in Base Ten) and Binary System Numbers (Base Two and Computer Representation):
1. Integer -> Binary
2. Decimal -> Binary
3. Binary -> Integer
4. Binary -> Decimal