-0.000 000 000 000 087 571 88 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 087 571 88(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 000 087 571 88(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 000 087 571 88| = 0.000 000 000 000 087 571 88


2. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

3. 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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 000 087 571 88.

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) 0.000 000 000 000 087 571 88 × 2 = 0 + 0.000 000 000 000 175 143 76;
  • 2) 0.000 000 000 000 175 143 76 × 2 = 0 + 0.000 000 000 000 350 287 52;
  • 3) 0.000 000 000 000 350 287 52 × 2 = 0 + 0.000 000 000 000 700 575 04;
  • 4) 0.000 000 000 000 700 575 04 × 2 = 0 + 0.000 000 000 001 401 150 08;
  • 5) 0.000 000 000 001 401 150 08 × 2 = 0 + 0.000 000 000 002 802 300 16;
  • 6) 0.000 000 000 002 802 300 16 × 2 = 0 + 0.000 000 000 005 604 600 32;
  • 7) 0.000 000 000 005 604 600 32 × 2 = 0 + 0.000 000 000 011 209 200 64;
  • 8) 0.000 000 000 011 209 200 64 × 2 = 0 + 0.000 000 000 022 418 401 28;
  • 9) 0.000 000 000 022 418 401 28 × 2 = 0 + 0.000 000 000 044 836 802 56;
  • 10) 0.000 000 000 044 836 802 56 × 2 = 0 + 0.000 000 000 089 673 605 12;
  • 11) 0.000 000 000 089 673 605 12 × 2 = 0 + 0.000 000 000 179 347 210 24;
  • 12) 0.000 000 000 179 347 210 24 × 2 = 0 + 0.000 000 000 358 694 420 48;
  • 13) 0.000 000 000 358 694 420 48 × 2 = 0 + 0.000 000 000 717 388 840 96;
  • 14) 0.000 000 000 717 388 840 96 × 2 = 0 + 0.000 000 001 434 777 681 92;
  • 15) 0.000 000 001 434 777 681 92 × 2 = 0 + 0.000 000 002 869 555 363 84;
  • 16) 0.000 000 002 869 555 363 84 × 2 = 0 + 0.000 000 005 739 110 727 68;
  • 17) 0.000 000 005 739 110 727 68 × 2 = 0 + 0.000 000 011 478 221 455 36;
  • 18) 0.000 000 011 478 221 455 36 × 2 = 0 + 0.000 000 022 956 442 910 72;
  • 19) 0.000 000 022 956 442 910 72 × 2 = 0 + 0.000 000 045 912 885 821 44;
  • 20) 0.000 000 045 912 885 821 44 × 2 = 0 + 0.000 000 091 825 771 642 88;
  • 21) 0.000 000 091 825 771 642 88 × 2 = 0 + 0.000 000 183 651 543 285 76;
  • 22) 0.000 000 183 651 543 285 76 × 2 = 0 + 0.000 000 367 303 086 571 52;
  • 23) 0.000 000 367 303 086 571 52 × 2 = 0 + 0.000 000 734 606 173 143 04;
  • 24) 0.000 000 734 606 173 143 04 × 2 = 0 + 0.000 001 469 212 346 286 08;
  • 25) 0.000 001 469 212 346 286 08 × 2 = 0 + 0.000 002 938 424 692 572 16;
  • 26) 0.000 002 938 424 692 572 16 × 2 = 0 + 0.000 005 876 849 385 144 32;
  • 27) 0.000 005 876 849 385 144 32 × 2 = 0 + 0.000 011 753 698 770 288 64;
  • 28) 0.000 011 753 698 770 288 64 × 2 = 0 + 0.000 023 507 397 540 577 28;
  • 29) 0.000 023 507 397 540 577 28 × 2 = 0 + 0.000 047 014 795 081 154 56;
  • 30) 0.000 047 014 795 081 154 56 × 2 = 0 + 0.000 094 029 590 162 309 12;
  • 31) 0.000 094 029 590 162 309 12 × 2 = 0 + 0.000 188 059 180 324 618 24;
  • 32) 0.000 188 059 180 324 618 24 × 2 = 0 + 0.000 376 118 360 649 236 48;
  • 33) 0.000 376 118 360 649 236 48 × 2 = 0 + 0.000 752 236 721 298 472 96;
  • 34) 0.000 752 236 721 298 472 96 × 2 = 0 + 0.001 504 473 442 596 945 92;
  • 35) 0.001 504 473 442 596 945 92 × 2 = 0 + 0.003 008 946 885 193 891 84;
  • 36) 0.003 008 946 885 193 891 84 × 2 = 0 + 0.006 017 893 770 387 783 68;
  • 37) 0.006 017 893 770 387 783 68 × 2 = 0 + 0.012 035 787 540 775 567 36;
  • 38) 0.012 035 787 540 775 567 36 × 2 = 0 + 0.024 071 575 081 551 134 72;
  • 39) 0.024 071 575 081 551 134 72 × 2 = 0 + 0.048 143 150 163 102 269 44;
  • 40) 0.048 143 150 163 102 269 44 × 2 = 0 + 0.096 286 300 326 204 538 88;
  • 41) 0.096 286 300 326 204 538 88 × 2 = 0 + 0.192 572 600 652 409 077 76;
  • 42) 0.192 572 600 652 409 077 76 × 2 = 0 + 0.385 145 201 304 818 155 52;
  • 43) 0.385 145 201 304 818 155 52 × 2 = 0 + 0.770 290 402 609 636 311 04;
  • 44) 0.770 290 402 609 636 311 04 × 2 = 1 + 0.540 580 805 219 272 622 08;
  • 45) 0.540 580 805 219 272 622 08 × 2 = 1 + 0.081 161 610 438 545 244 16;
  • 46) 0.081 161 610 438 545 244 16 × 2 = 0 + 0.162 323 220 877 090 488 32;
  • 47) 0.162 323 220 877 090 488 32 × 2 = 0 + 0.324 646 441 754 180 976 64;
  • 48) 0.324 646 441 754 180 976 64 × 2 = 0 + 0.649 292 883 508 361 953 28;
  • 49) 0.649 292 883 508 361 953 28 × 2 = 1 + 0.298 585 767 016 723 906 56;
  • 50) 0.298 585 767 016 723 906 56 × 2 = 0 + 0.597 171 534 033 447 813 12;
  • 51) 0.597 171 534 033 447 813 12 × 2 = 1 + 0.194 343 068 066 895 626 24;
  • 52) 0.194 343 068 066 895 626 24 × 2 = 0 + 0.388 686 136 133 791 252 48;
  • 53) 0.388 686 136 133 791 252 48 × 2 = 0 + 0.777 372 272 267 582 504 96;
  • 54) 0.777 372 272 267 582 504 96 × 2 = 1 + 0.554 744 544 535 165 009 92;
  • 55) 0.554 744 544 535 165 009 92 × 2 = 1 + 0.109 489 089 070 330 019 84;
  • 56) 0.109 489 089 070 330 019 84 × 2 = 0 + 0.218 978 178 140 660 039 68;
  • 57) 0.218 978 178 140 660 039 68 × 2 = 0 + 0.437 956 356 281 320 079 36;
  • 58) 0.437 956 356 281 320 079 36 × 2 = 0 + 0.875 912 712 562 640 158 72;
  • 59) 0.875 912 712 562 640 158 72 × 2 = 1 + 0.751 825 425 125 280 317 44;
  • 60) 0.751 825 425 125 280 317 44 × 2 = 1 + 0.503 650 850 250 560 634 88;
  • 61) 0.503 650 850 250 560 634 88 × 2 = 1 + 0.007 301 700 501 121 269 76;
  • 62) 0.007 301 700 501 121 269 76 × 2 = 0 + 0.014 603 401 002 242 539 52;
  • 63) 0.014 603 401 002 242 539 52 × 2 = 0 + 0.029 206 802 004 485 079 04;
  • 64) 0.029 206 802 004 485 079 04 × 2 = 0 + 0.058 413 604 008 970 158 08;
  • 65) 0.058 413 604 008 970 158 08 × 2 = 0 + 0.116 827 208 017 940 316 16;
  • 66) 0.116 827 208 017 940 316 16 × 2 = 0 + 0.233 654 416 035 880 632 32;
  • 67) 0.233 654 416 035 880 632 32 × 2 = 0 + 0.467 308 832 071 761 264 64;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


5. 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:


0.000 000 000 000 087 571 88(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1000 1010 0110 0011 1000 000(2)

6. Positive number before normalization:

0.000 000 000 000 087 571 88(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1000 1010 0110 0011 1000 000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 44 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 087 571 88(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1000 1010 0110 0011 1000 000(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1000 1010 0110 0011 1000 000(2) × 20 =

1.1000 1010 0110 0011 1000 000(2) × 2-44


8. 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 1 (a negative number)

Exponent (unadjusted): -44

Mantissa (not normalized):
1.1000 1010 0110 0011 1000 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-44 + 2(8-1) - 1 =

(-44 + 127)(10) =

83(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 83 ÷ 2 = 41 + 1;
  • 41 ÷ 2 = 20 + 1;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

83(10) =

0101 0011(2)


12. 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 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 0101 0011 0001 1100 0000 =

100 0101 0011 0001 1100 0000


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

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

Exponent (8 bits) =
0101 0011

Mantissa (23 bits) =
100 0101 0011 0001 1100 0000


Decimal number -0.000 000 000 000 087 571 88 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0101 0011 - 100 0101 0011 0001 1100 0000

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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