32bit IEEE 754: Decimal -> Single Precision Floating Point Binary: -0.000 000 000 000 000 001 1 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number -0.000 000 000 000 000 001 1(10) converted and written in 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:

|-0.000 000 000 000 000 001 1| = 0.000 000 000 000 000 001 1

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: 0.000 000 000 000 000 001 1.

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) 0.000 000 000 000 000 001 1 × 2 = 0 + 0.000 000 000 000 000 002 2;
  • 2) 0.000 000 000 000 000 002 2 × 2 = 0 + 0.000 000 000 000 000 004 4;
  • 3) 0.000 000 000 000 000 004 4 × 2 = 0 + 0.000 000 000 000 000 008 8;
  • 4) 0.000 000 000 000 000 008 8 × 2 = 0 + 0.000 000 000 000 000 017 6;
  • 5) 0.000 000 000 000 000 017 6 × 2 = 0 + 0.000 000 000 000 000 035 2;
  • 6) 0.000 000 000 000 000 035 2 × 2 = 0 + 0.000 000 000 000 000 070 4;
  • 7) 0.000 000 000 000 000 070 4 × 2 = 0 + 0.000 000 000 000 000 140 8;
  • 8) 0.000 000 000 000 000 140 8 × 2 = 0 + 0.000 000 000 000 000 281 6;
  • 9) 0.000 000 000 000 000 281 6 × 2 = 0 + 0.000 000 000 000 000 563 2;
  • 10) 0.000 000 000 000 000 563 2 × 2 = 0 + 0.000 000 000 000 001 126 4;
  • 11) 0.000 000 000 000 001 126 4 × 2 = 0 + 0.000 000 000 000 002 252 8;
  • 12) 0.000 000 000 000 002 252 8 × 2 = 0 + 0.000 000 000 000 004 505 6;
  • 13) 0.000 000 000 000 004 505 6 × 2 = 0 + 0.000 000 000 000 009 011 2;
  • 14) 0.000 000 000 000 009 011 2 × 2 = 0 + 0.000 000 000 000 018 022 4;
  • 15) 0.000 000 000 000 018 022 4 × 2 = 0 + 0.000 000 000 000 036 044 8;
  • 16) 0.000 000 000 000 036 044 8 × 2 = 0 + 0.000 000 000 000 072 089 6;
  • 17) 0.000 000 000 000 072 089 6 × 2 = 0 + 0.000 000 000 000 144 179 2;
  • 18) 0.000 000 000 000 144 179 2 × 2 = 0 + 0.000 000 000 000 288 358 4;
  • 19) 0.000 000 000 000 288 358 4 × 2 = 0 + 0.000 000 000 000 576 716 8;
  • 20) 0.000 000 000 000 576 716 8 × 2 = 0 + 0.000 000 000 001 153 433 6;
  • 21) 0.000 000 000 001 153 433 6 × 2 = 0 + 0.000 000 000 002 306 867 2;
  • 22) 0.000 000 000 002 306 867 2 × 2 = 0 + 0.000 000 000 004 613 734 4;
  • 23) 0.000 000 000 004 613 734 4 × 2 = 0 + 0.000 000 000 009 227 468 8;
  • 24) 0.000 000 000 009 227 468 8 × 2 = 0 + 0.000 000 000 018 454 937 6;
  • 25) 0.000 000 000 018 454 937 6 × 2 = 0 + 0.000 000 000 036 909 875 2;
  • 26) 0.000 000 000 036 909 875 2 × 2 = 0 + 0.000 000 000 073 819 750 4;
  • 27) 0.000 000 000 073 819 750 4 × 2 = 0 + 0.000 000 000 147 639 500 8;
  • 28) 0.000 000 000 147 639 500 8 × 2 = 0 + 0.000 000 000 295 279 001 6;
  • 29) 0.000 000 000 295 279 001 6 × 2 = 0 + 0.000 000 000 590 558 003 2;
  • 30) 0.000 000 000 590 558 003 2 × 2 = 0 + 0.000 000 001 181 116 006 4;
  • 31) 0.000 000 001 181 116 006 4 × 2 = 0 + 0.000 000 002 362 232 012 8;
  • 32) 0.000 000 002 362 232 012 8 × 2 = 0 + 0.000 000 004 724 464 025 6;
  • 33) 0.000 000 004 724 464 025 6 × 2 = 0 + 0.000 000 009 448 928 051 2;
  • 34) 0.000 000 009 448 928 051 2 × 2 = 0 + 0.000 000 018 897 856 102 4;
  • 35) 0.000 000 018 897 856 102 4 × 2 = 0 + 0.000 000 037 795 712 204 8;
  • 36) 0.000 000 037 795 712 204 8 × 2 = 0 + 0.000 000 075 591 424 409 6;
  • 37) 0.000 000 075 591 424 409 6 × 2 = 0 + 0.000 000 151 182 848 819 2;
  • 38) 0.000 000 151 182 848 819 2 × 2 = 0 + 0.000 000 302 365 697 638 4;
  • 39) 0.000 000 302 365 697 638 4 × 2 = 0 + 0.000 000 604 731 395 276 8;
  • 40) 0.000 000 604 731 395 276 8 × 2 = 0 + 0.000 001 209 462 790 553 6;
  • 41) 0.000 001 209 462 790 553 6 × 2 = 0 + 0.000 002 418 925 581 107 2;
  • 42) 0.000 002 418 925 581 107 2 × 2 = 0 + 0.000 004 837 851 162 214 4;
  • 43) 0.000 004 837 851 162 214 4 × 2 = 0 + 0.000 009 675 702 324 428 8;
  • 44) 0.000 009 675 702 324 428 8 × 2 = 0 + 0.000 019 351 404 648 857 6;
  • 45) 0.000 019 351 404 648 857 6 × 2 = 0 + 0.000 038 702 809 297 715 2;
  • 46) 0.000 038 702 809 297 715 2 × 2 = 0 + 0.000 077 405 618 595 430 4;
  • 47) 0.000 077 405 618 595 430 4 × 2 = 0 + 0.000 154 811 237 190 860 8;
  • 48) 0.000 154 811 237 190 860 8 × 2 = 0 + 0.000 309 622 474 381 721 6;
  • 49) 0.000 309 622 474 381 721 6 × 2 = 0 + 0.000 619 244 948 763 443 2;
  • 50) 0.000 619 244 948 763 443 2 × 2 = 0 + 0.001 238 489 897 526 886 4;
  • 51) 0.001 238 489 897 526 886 4 × 2 = 0 + 0.002 476 979 795 053 772 8;
  • 52) 0.002 476 979 795 053 772 8 × 2 = 0 + 0.004 953 959 590 107 545 6;
  • 53) 0.004 953 959 590 107 545 6 × 2 = 0 + 0.009 907 919 180 215 091 2;
  • 54) 0.009 907 919 180 215 091 2 × 2 = 0 + 0.019 815 838 360 430 182 4;
  • 55) 0.019 815 838 360 430 182 4 × 2 = 0 + 0.039 631 676 720 860 364 8;
  • 56) 0.039 631 676 720 860 364 8 × 2 = 0 + 0.079 263 353 441 720 729 6;
  • 57) 0.079 263 353 441 720 729 6 × 2 = 0 + 0.158 526 706 883 441 459 2;
  • 58) 0.158 526 706 883 441 459 2 × 2 = 0 + 0.317 053 413 766 882 918 4;
  • 59) 0.317 053 413 766 882 918 4 × 2 = 0 + 0.634 106 827 533 765 836 8;
  • 60) 0.634 106 827 533 765 836 8 × 2 = 1 + 0.268 213 655 067 531 673 6;
  • 61) 0.268 213 655 067 531 673 6 × 2 = 0 + 0.536 427 310 135 063 347 2;
  • 62) 0.536 427 310 135 063 347 2 × 2 = 1 + 0.072 854 620 270 126 694 4;
  • 63) 0.072 854 620 270 126 694 4 × 2 = 0 + 0.145 709 240 540 253 388 8;
  • 64) 0.145 709 240 540 253 388 8 × 2 = 0 + 0.291 418 481 080 506 777 6;
  • 65) 0.291 418 481 080 506 777 6 × 2 = 0 + 0.582 836 962 161 013 555 2;
  • 66) 0.582 836 962 161 013 555 2 × 2 = 1 + 0.165 673 924 322 027 110 4;
  • 67) 0.165 673 924 322 027 110 4 × 2 = 0 + 0.331 347 848 644 054 220 8;
  • 68) 0.331 347 848 644 054 220 8 × 2 = 0 + 0.662 695 697 288 108 441 6;
  • 69) 0.662 695 697 288 108 441 6 × 2 = 1 + 0.325 391 394 576 216 883 2;
  • 70) 0.325 391 394 576 216 883 2 × 2 = 0 + 0.650 782 789 152 433 766 4;
  • 71) 0.650 782 789 152 433 766 4 × 2 = 1 + 0.301 565 578 304 867 532 8;
  • 72) 0.301 565 578 304 867 532 8 × 2 = 0 + 0.603 131 156 609 735 065 6;
  • 73) 0.603 131 156 609 735 065 6 × 2 = 1 + 0.206 262 313 219 470 131 2;
  • 74) 0.206 262 313 219 470 131 2 × 2 = 0 + 0.412 524 626 438 940 262 4;
  • 75) 0.412 524 626 438 940 262 4 × 2 = 0 + 0.825 049 252 877 880 524 8;
  • 76) 0.825 049 252 877 880 524 8 × 2 = 1 + 0.650 098 505 755 761 049 6;
  • 77) 0.650 098 505 755 761 049 6 × 2 = 1 + 0.300 197 011 511 522 099 2;
  • 78) 0.300 197 011 511 522 099 2 × 2 = 0 + 0.600 394 023 023 044 198 4;
  • 79) 0.600 394 023 023 044 198 4 × 2 = 1 + 0.200 788 046 046 088 396 8;
  • 80) 0.200 788 046 046 088 396 8 × 2 = 0 + 0.401 576 092 092 176 793 6;
  • 81) 0.401 576 092 092 176 793 6 × 2 = 0 + 0.803 152 184 184 353 587 2;
  • 82) 0.803 152 184 184 353 587 2 × 2 = 1 + 0.606 304 368 368 707 174 4;
  • 83) 0.606 304 368 368 707 174 4 × 2 = 1 + 0.212 608 736 737 414 348 8;

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...)


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:


0.000 000 000 000 000 001 1(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0100 0100 1010 1001 1010 011(2)


6. Positive number before normalization:

0.000 000 000 000 000 001 1(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0100 0100 1010 1001 1010 011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 60 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 001 1(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0100 0100 1010 1001 1010 011(2) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0100 0100 1010 1001 1010 011(2) × 20 =


1.0100 0100 1010 1001 1010 011(2) × 2-60


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): -60


Mantissa (not normalized):
1.0100 0100 1010 1001 1010 011


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


-60 + 2(8-1) - 1 =


(-60 + 127)(10) =


67(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;
  • 67 ÷ 2 = 33 + 1;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 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) =


67(10) =


0100 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. 010 0010 0101 0100 1101 0011 =


010 0010 0101 0100 1101 0011


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) =
0100 0011


Mantissa (23 bits) =
010 0010 0101 0100 1101 0011


The base ten decimal number -0.000 000 000 000 000 001 1 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
1 - 0100 0011 - 010 0010 0101 0100 1101 0011

(32 bits IEEE 754)
  • Sign (1 bit):

    • 1

      31
  • Exponent (8 bits):

    • 0

      30
    • 1

      29
    • 0

      28
    • 0

      27
    • 0

      26
    • 0

      25
    • 1

      24
    • 1

      23
  • Mantissa (23 bits):

    • 0

      22
    • 1

      21
    • 0

      20
    • 0

      19
    • 0

      18
    • 1

      17
    • 0

      16
    • 0

      15
    • 1

      14
    • 0

      13
    • 1

      12
    • 0

      11
    • 1

      10
    • 0

      9
    • 0

      8
    • 1

      7
    • 1

      6
    • 0

      5
    • 1

      4
    • 0

      3
    • 0

      2
    • 1

      1
    • 1

      0

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All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

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