Base ten decimal number 0.000 000 000 000 000 000 000 01 converted to 32 bit single precision IEEE 754 binary floating point standard

How to convert the decimal number 0.000 000 000 000 000 000 000 01(10)
to
32 bit single precision IEEE 754 binary floating point
(1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. First, convert to binary (base 2) the integer part: 0. Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:

  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

2. Construct the base 2 representation of the integer part of the number, by taking all the remainders starting from the bottom of the list constructed above:

0(10) =


0(2)

3. Convert to binary (base 2) the fractional part: 0.000 000 000 000 000 000 000 01. Multiply it 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) 0.000 000 000 000 000 000 000 01 × 2 = 0 + 0.000 000 000 000 000 000 000 02;
  • 2) 0.000 000 000 000 000 000 000 02 × 2 = 0 + 0.000 000 000 000 000 000 000 04;
  • 3) 0.000 000 000 000 000 000 000 04 × 2 = 0 + 0.000 000 000 000 000 000 000 08;
  • 4) 0.000 000 000 000 000 000 000 08 × 2 = 0 + 0.000 000 000 000 000 000 000 16;
  • 5) 0.000 000 000 000 000 000 000 16 × 2 = 0 + 0.000 000 000 000 000 000 000 32;
  • 6) 0.000 000 000 000 000 000 000 32 × 2 = 0 + 0.000 000 000 000 000 000 000 64;
  • 7) 0.000 000 000 000 000 000 000 64 × 2 = 0 + 0.000 000 000 000 000 000 001 28;
  • 8) 0.000 000 000 000 000 000 001 28 × 2 = 0 + 0.000 000 000 000 000 000 002 56;
  • 9) 0.000 000 000 000 000 000 002 56 × 2 = 0 + 0.000 000 000 000 000 000 005 12;
  • 10) 0.000 000 000 000 000 000 005 12 × 2 = 0 + 0.000 000 000 000 000 000 010 24;
  • 11) 0.000 000 000 000 000 000 010 24 × 2 = 0 + 0.000 000 000 000 000 000 020 48;
  • 12) 0.000 000 000 000 000 000 020 48 × 2 = 0 + 0.000 000 000 000 000 000 040 96;
  • 13) 0.000 000 000 000 000 000 040 96 × 2 = 0 + 0.000 000 000 000 000 000 081 92;
  • 14) 0.000 000 000 000 000 000 081 92 × 2 = 0 + 0.000 000 000 000 000 000 163 84;
  • 15) 0.000 000 000 000 000 000 163 84 × 2 = 0 + 0.000 000 000 000 000 000 327 68;
  • 16) 0.000 000 000 000 000 000 327 68 × 2 = 0 + 0.000 000 000 000 000 000 655 36;
  • 17) 0.000 000 000 000 000 000 655 36 × 2 = 0 + 0.000 000 000 000 000 001 310 72;
  • 18) 0.000 000 000 000 000 001 310 72 × 2 = 0 + 0.000 000 000 000 000 002 621 44;
  • 19) 0.000 000 000 000 000 002 621 44 × 2 = 0 + 0.000 000 000 000 000 005 242 88;
  • 20) 0.000 000 000 000 000 005 242 88 × 2 = 0 + 0.000 000 000 000 000 010 485 76;
  • 21) 0.000 000 000 000 000 010 485 76 × 2 = 0 + 0.000 000 000 000 000 020 971 52;
  • 22) 0.000 000 000 000 000 020 971 52 × 2 = 0 + 0.000 000 000 000 000 041 943 04;
  • 23) 0.000 000 000 000 000 041 943 04 × 2 = 0 + 0.000 000 000 000 000 083 886 08;
  • 24) 0.000 000 000 000 000 083 886 08 × 2 = 0 + 0.000 000 000 000 000 167 772 16;
  • 25) 0.000 000 000 000 000 167 772 16 × 2 = 0 + 0.000 000 000 000 000 335 544 32;
  • 26) 0.000 000 000 000 000 335 544 32 × 2 = 0 + 0.000 000 000 000 000 671 088 64;
  • 27) 0.000 000 000 000 000 671 088 64 × 2 = 0 + 0.000 000 000 000 001 342 177 28;
  • 28) 0.000 000 000 000 001 342 177 28 × 2 = 0 + 0.000 000 000 000 002 684 354 56;
  • 29) 0.000 000 000 000 002 684 354 56 × 2 = 0 + 0.000 000 000 000 005 368 709 12;
  • 30) 0.000 000 000 000 005 368 709 12 × 2 = 0 + 0.000 000 000 000 010 737 418 24;
  • 31) 0.000 000 000 000 010 737 418 24 × 2 = 0 + 0.000 000 000 000 021 474 836 48;
  • 32) 0.000 000 000 000 021 474 836 48 × 2 = 0 + 0.000 000 000 000 042 949 672 96;
  • 33) 0.000 000 000 000 042 949 672 96 × 2 = 0 + 0.000 000 000 000 085 899 345 92;
  • 34) 0.000 000 000 000 085 899 345 92 × 2 = 0 + 0.000 000 000 000 171 798 691 84;
  • 35) 0.000 000 000 000 171 798 691 84 × 2 = 0 + 0.000 000 000 000 343 597 383 68;
  • 36) 0.000 000 000 000 343 597 383 68 × 2 = 0 + 0.000 000 000 000 687 194 767 36;
  • 37) 0.000 000 000 000 687 194 767 36 × 2 = 0 + 0.000 000 000 001 374 389 534 72;
  • 38) 0.000 000 000 001 374 389 534 72 × 2 = 0 + 0.000 000 000 002 748 779 069 44;
  • 39) 0.000 000 000 002 748 779 069 44 × 2 = 0 + 0.000 000 000 005 497 558 138 88;
  • 40) 0.000 000 000 005 497 558 138 88 × 2 = 0 + 0.000 000 000 010 995 116 277 76;
  • 41) 0.000 000 000 010 995 116 277 76 × 2 = 0 + 0.000 000 000 021 990 232 555 52;
  • 42) 0.000 000 000 021 990 232 555 52 × 2 = 0 + 0.000 000 000 043 980 465 111 04;
  • 43) 0.000 000 000 043 980 465 111 04 × 2 = 0 + 0.000 000 000 087 960 930 222 08;
  • 44) 0.000 000 000 087 960 930 222 08 × 2 = 0 + 0.000 000 000 175 921 860 444 16;
  • 45) 0.000 000 000 175 921 860 444 16 × 2 = 0 + 0.000 000 000 351 843 720 888 32;
  • 46) 0.000 000 000 351 843 720 888 32 × 2 = 0 + 0.000 000 000 703 687 441 776 64;
  • 47) 0.000 000 000 703 687 441 776 64 × 2 = 0 + 0.000 000 001 407 374 883 553 28;
  • 48) 0.000 000 001 407 374 883 553 28 × 2 = 0 + 0.000 000 002 814 749 767 106 56;
  • 49) 0.000 000 002 814 749 767 106 56 × 2 = 0 + 0.000 000 005 629 499 534 213 12;
  • 50) 0.000 000 005 629 499 534 213 12 × 2 = 0 + 0.000 000 011 258 999 068 426 24;
  • 51) 0.000 000 011 258 999 068 426 24 × 2 = 0 + 0.000 000 022 517 998 136 852 48;
  • 52) 0.000 000 022 517 998 136 852 48 × 2 = 0 + 0.000 000 045 035 996 273 704 96;
  • 53) 0.000 000 045 035 996 273 704 96 × 2 = 0 + 0.000 000 090 071 992 547 409 92;
  • 54) 0.000 000 090 071 992 547 409 92 × 2 = 0 + 0.000 000 180 143 985 094 819 84;
  • 55) 0.000 000 180 143 985 094 819 84 × 2 = 0 + 0.000 000 360 287 970 189 639 68;
  • 56) 0.000 000 360 287 970 189 639 68 × 2 = 0 + 0.000 000 720 575 940 379 279 36;
  • 57) 0.000 000 720 575 940 379 279 36 × 2 = 0 + 0.000 001 441 151 880 758 558 72;
  • 58) 0.000 001 441 151 880 758 558 72 × 2 = 0 + 0.000 002 882 303 761 517 117 44;
  • 59) 0.000 002 882 303 761 517 117 44 × 2 = 0 + 0.000 005 764 607 523 034 234 88;
  • 60) 0.000 005 764 607 523 034 234 88 × 2 = 0 + 0.000 011 529 215 046 068 469 76;
  • 61) 0.000 011 529 215 046 068 469 76 × 2 = 0 + 0.000 023 058 430 092 136 939 52;
  • 62) 0.000 023 058 430 092 136 939 52 × 2 = 0 + 0.000 046 116 860 184 273 879 04;
  • 63) 0.000 046 116 860 184 273 879 04 × 2 = 0 + 0.000 092 233 720 368 547 758 08;
  • 64) 0.000 092 233 720 368 547 758 08 × 2 = 0 + 0.000 184 467 440 737 095 516 16;
  • 65) 0.000 184 467 440 737 095 516 16 × 2 = 0 + 0.000 368 934 881 474 191 032 32;
  • 66) 0.000 368 934 881 474 191 032 32 × 2 = 0 + 0.000 737 869 762 948 382 064 64;
  • 67) 0.000 737 869 762 948 382 064 64 × 2 = 0 + 0.001 475 739 525 896 764 129 28;
  • 68) 0.001 475 739 525 896 764 129 28 × 2 = 0 + 0.002 951 479 051 793 528 258 56;
  • 69) 0.002 951 479 051 793 528 258 56 × 2 = 0 + 0.005 902 958 103 587 056 517 12;
  • 70) 0.005 902 958 103 587 056 517 12 × 2 = 0 + 0.011 805 916 207 174 113 034 24;
  • 71) 0.011 805 916 207 174 113 034 24 × 2 = 0 + 0.023 611 832 414 348 226 068 48;
  • 72) 0.023 611 832 414 348 226 068 48 × 2 = 0 + 0.047 223 664 828 696 452 136 96;
  • 73) 0.047 223 664 828 696 452 136 96 × 2 = 0 + 0.094 447 329 657 392 904 273 92;
  • 74) 0.094 447 329 657 392 904 273 92 × 2 = 0 + 0.188 894 659 314 785 808 547 84;
  • 75) 0.188 894 659 314 785 808 547 84 × 2 = 0 + 0.377 789 318 629 571 617 095 68;
  • 76) 0.377 789 318 629 571 617 095 68 × 2 = 0 + 0.755 578 637 259 143 234 191 36;
  • 77) 0.755 578 637 259 143 234 191 36 × 2 = 1 + 0.511 157 274 518 286 468 382 72;
  • 78) 0.511 157 274 518 286 468 382 72 × 2 = 1 + 0.022 314 549 036 572 936 765 44;
  • 79) 0.022 314 549 036 572 936 765 44 × 2 = 0 + 0.044 629 098 073 145 873 530 88;
  • 80) 0.044 629 098 073 145 873 530 88 × 2 = 0 + 0.089 258 196 146 291 747 061 76;
  • 81) 0.089 258 196 146 291 747 061 76 × 2 = 0 + 0.178 516 392 292 583 494 123 52;
  • 82) 0.178 516 392 292 583 494 123 52 × 2 = 0 + 0.357 032 784 585 166 988 247 04;
  • 83) 0.357 032 784 585 166 988 247 04 × 2 = 0 + 0.714 065 569 170 333 976 494 08;
  • 84) 0.714 065 569 170 333 976 494 08 × 2 = 1 + 0.428 131 138 340 667 952 988 16;
  • 85) 0.428 131 138 340 667 952 988 16 × 2 = 0 + 0.856 262 276 681 335 905 976 32;
  • 86) 0.856 262 276 681 335 905 976 32 × 2 = 1 + 0.712 524 553 362 671 811 952 64;
  • 87) 0.712 524 553 362 671 811 952 64 × 2 = 1 + 0.425 049 106 725 343 623 905 28;
  • 88) 0.425 049 106 725 343 623 905 28 × 2 = 0 + 0.850 098 213 450 687 247 810 56;
  • 89) 0.850 098 213 450 687 247 810 56 × 2 = 1 + 0.700 196 426 901 374 495 621 12;
  • 90) 0.700 196 426 901 374 495 621 12 × 2 = 1 + 0.400 392 853 802 748 991 242 24;
  • 91) 0.400 392 853 802 748 991 242 24 × 2 = 0 + 0.800 785 707 605 497 982 484 48;
  • 92) 0.800 785 707 605 497 982 484 48 × 2 = 1 + 0.601 571 415 210 995 964 968 96;
  • 93) 0.601 571 415 210 995 964 968 96 × 2 = 1 + 0.203 142 830 421 991 929 937 92;
  • 94) 0.203 142 830 421 991 929 937 92 × 2 = 0 + 0.406 285 660 843 983 859 875 84;
  • 95) 0.406 285 660 843 983 859 875 84 × 2 = 0 + 0.812 571 321 687 967 719 751 68;
  • 96) 0.812 571 321 687 967 719 751 68 × 2 = 1 + 0.625 142 643 375 935 439 503 36;
  • 97) 0.625 142 643 375 935 439 503 36 × 2 = 1 + 0.250 285 286 751 870 879 006 72;
  • 98) 0.250 285 286 751 870 879 006 72 × 2 = 0 + 0.500 570 573 503 741 758 013 44;
  • 99) 0.500 570 573 503 741 758 013 44 × 2 = 1 + 0.001 141 147 007 483 516 026 88;
  • 100) 0.001 141 147 007 483 516 026 88 × 2 = 0 + 0.002 282 294 014 967 032 053 76;

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, by taking all the integer parts of the multiplying operations, starting from the top of the constructed list above:

0.000 000 000 000 000 000 000 01(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 0001 0110 1101 1001 1010(2)

Positive number before normalization:

0.000 000 000 000 000 000 000 01(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 0001 0110 1101 1001 1010(2)

5. Normalize the binary representation of the number, shifting the decimal mark 77 positions to the right so that only one non zero digit remains to the left of it:

0.000 000 000 000 000 000 000 01(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 0001 0110 1101 1001 1010(2) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 0001 0110 1101 1001 1010(2) × 20 =


1.1000 0010 1101 1011 0011 010(2) × 2-77

Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign: 0 (a positive number)


Exponent (unadjusted): -77


Mantissa (not normalized): 1.1000 0010 1101 1011 0011 010

6. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2:

Exponent (adjusted) =


Exponent (unadjusted) + 2(8-1) - 1 =


-77 + 2(8-1) - 1 =


(-77 + 127)(10) =


50(10)


  • division = quotient + remainder;
  • 50 ÷ 2 = 25 + 0;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

Exponent (adjusted) =


50(10) =


0011 0010(2)

7. Normalize mantissa, remove the leading (the leftmost) bit, since it's allways 1 (and the decimal point, if the case) then adjust its length to 23 bits, only if necessary (not the case here):

Mantissa (normalized) =


1. 100 0001 0110 1101 1001 1010 =


100 0001 0110 1101 1001 1010

Conclusion:

The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)


Exponent (8 bits) =
0011 0010


Mantissa (23 bits) =
100 0001 0110 1101 1001 1010

Number 0.000 000 000 000 000 000 000 01, a decimal, converted from decimal system (base 10)
to
32 bit single precision IEEE 754 binary floating point:


0 - 0011 0010 - 100 0001 0110 1101 1001 1010

(32 bits IEEE 754)
  • Sign (1 bit):

    • 0

      31
  • Exponent (8 bits):

    • 0

      30
    • 0

      29
    • 1

      28
    • 1

      27
    • 0

      26
    • 0

      25
    • 1

      24
    • 0

      23
  • Mantissa (23 bits):

    • 1

      22
    • 0

      21
    • 0

      20
    • 0

      19
    • 0

      18
    • 0

      17
    • 1

      16
    • 0

      15
    • 1

      14
    • 1

      13
    • 0

      12
    • 1

      11
    • 1

      10
    • 0

      9
    • 1

      8
    • 1

      7
    • 0

      6
    • 0

      5
    • 1

      4
    • 1

      3
    • 0

      2
    • 1

      1
    • 0

      0

Convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

A number in 32 bit single 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 (8 bits), mantissa (23 bits)

Latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation

How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single 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 base ten 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 of the previous dividing 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 previous multiplying operations, starting from the top of the constructed list 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, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits 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 -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 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:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. 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) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) 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:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =


    1 - 1000 0011 - 100 1010 1100 0110 1010 0111