-0.000 000 000 741 959 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 741 959(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
-0.000 000 000 741 959(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:

|-0.000 000 000 741 959| = 0.000 000 000 741 959


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: 0.000 000 000 741 959.

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) 0.000 000 000 741 959 × 2 = 0 + 0.000 000 001 483 918;
  • 2) 0.000 000 001 483 918 × 2 = 0 + 0.000 000 002 967 836;
  • 3) 0.000 000 002 967 836 × 2 = 0 + 0.000 000 005 935 672;
  • 4) 0.000 000 005 935 672 × 2 = 0 + 0.000 000 011 871 344;
  • 5) 0.000 000 011 871 344 × 2 = 0 + 0.000 000 023 742 688;
  • 6) 0.000 000 023 742 688 × 2 = 0 + 0.000 000 047 485 376;
  • 7) 0.000 000 047 485 376 × 2 = 0 + 0.000 000 094 970 752;
  • 8) 0.000 000 094 970 752 × 2 = 0 + 0.000 000 189 941 504;
  • 9) 0.000 000 189 941 504 × 2 = 0 + 0.000 000 379 883 008;
  • 10) 0.000 000 379 883 008 × 2 = 0 + 0.000 000 759 766 016;
  • 11) 0.000 000 759 766 016 × 2 = 0 + 0.000 001 519 532 032;
  • 12) 0.000 001 519 532 032 × 2 = 0 + 0.000 003 039 064 064;
  • 13) 0.000 003 039 064 064 × 2 = 0 + 0.000 006 078 128 128;
  • 14) 0.000 006 078 128 128 × 2 = 0 + 0.000 012 156 256 256;
  • 15) 0.000 012 156 256 256 × 2 = 0 + 0.000 024 312 512 512;
  • 16) 0.000 024 312 512 512 × 2 = 0 + 0.000 048 625 025 024;
  • 17) 0.000 048 625 025 024 × 2 = 0 + 0.000 097 250 050 048;
  • 18) 0.000 097 250 050 048 × 2 = 0 + 0.000 194 500 100 096;
  • 19) 0.000 194 500 100 096 × 2 = 0 + 0.000 389 000 200 192;
  • 20) 0.000 389 000 200 192 × 2 = 0 + 0.000 778 000 400 384;
  • 21) 0.000 778 000 400 384 × 2 = 0 + 0.001 556 000 800 768;
  • 22) 0.001 556 000 800 768 × 2 = 0 + 0.003 112 001 601 536;
  • 23) 0.003 112 001 601 536 × 2 = 0 + 0.006 224 003 203 072;
  • 24) 0.006 224 003 203 072 × 2 = 0 + 0.012 448 006 406 144;
  • 25) 0.012 448 006 406 144 × 2 = 0 + 0.024 896 012 812 288;
  • 26) 0.024 896 012 812 288 × 2 = 0 + 0.049 792 025 624 576;
  • 27) 0.049 792 025 624 576 × 2 = 0 + 0.099 584 051 249 152;
  • 28) 0.099 584 051 249 152 × 2 = 0 + 0.199 168 102 498 304;
  • 29) 0.199 168 102 498 304 × 2 = 0 + 0.398 336 204 996 608;
  • 30) 0.398 336 204 996 608 × 2 = 0 + 0.796 672 409 993 216;
  • 31) 0.796 672 409 993 216 × 2 = 1 + 0.593 344 819 986 432;
  • 32) 0.593 344 819 986 432 × 2 = 1 + 0.186 689 639 972 864;
  • 33) 0.186 689 639 972 864 × 2 = 0 + 0.373 379 279 945 728;
  • 34) 0.373 379 279 945 728 × 2 = 0 + 0.746 758 559 891 456;
  • 35) 0.746 758 559 891 456 × 2 = 1 + 0.493 517 119 782 912;
  • 36) 0.493 517 119 782 912 × 2 = 0 + 0.987 034 239 565 824;
  • 37) 0.987 034 239 565 824 × 2 = 1 + 0.974 068 479 131 648;
  • 38) 0.974 068 479 131 648 × 2 = 1 + 0.948 136 958 263 296;
  • 39) 0.948 136 958 263 296 × 2 = 1 + 0.896 273 916 526 592;
  • 40) 0.896 273 916 526 592 × 2 = 1 + 0.792 547 833 053 184;
  • 41) 0.792 547 833 053 184 × 2 = 1 + 0.585 095 666 106 368;
  • 42) 0.585 095 666 106 368 × 2 = 1 + 0.170 191 332 212 736;
  • 43) 0.170 191 332 212 736 × 2 = 0 + 0.340 382 664 425 472;
  • 44) 0.340 382 664 425 472 × 2 = 0 + 0.680 765 328 850 944;
  • 45) 0.680 765 328 850 944 × 2 = 1 + 0.361 530 657 701 888;
  • 46) 0.361 530 657 701 888 × 2 = 0 + 0.723 061 315 403 776;
  • 47) 0.723 061 315 403 776 × 2 = 1 + 0.446 122 630 807 552;
  • 48) 0.446 122 630 807 552 × 2 = 0 + 0.892 245 261 615 104;
  • 49) 0.892 245 261 615 104 × 2 = 1 + 0.784 490 523 230 208;
  • 50) 0.784 490 523 230 208 × 2 = 1 + 0.568 981 046 460 416;
  • 51) 0.568 981 046 460 416 × 2 = 1 + 0.137 962 092 920 832;
  • 52) 0.137 962 092 920 832 × 2 = 0 + 0.275 924 185 841 664;
  • 53) 0.275 924 185 841 664 × 2 = 0 + 0.551 848 371 683 328;
  • 54) 0.551 848 371 683 328 × 2 = 1 + 0.103 696 743 366 656;

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:


0.000 000 000 741 959(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1100 1010 1110 01(2)

6. Positive number before normalization:

0.000 000 000 741 959(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1100 1010 1110 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 741 959(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1100 1010 1110 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1100 1010 1110 01(2) × 20 =

1.1001 0111 1110 0101 0111 001(2) × 2-31


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): -31

Mantissa (not normalized):
1.1001 0111 1110 0101 0111 001


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

96(10) =

0110 0000(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 1011 1111 0010 1011 1001 =

100 1011 1111 0010 1011 1001


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) =
0110 0000

Mantissa (23 bits) =
100 1011 1111 0010 1011 1001


Decimal number -0.000 000 000 741 959 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 0010 1011 1001

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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