-0.000 000 000 742 094 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 094(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
-0.000 000 000 742 094(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:

|-0.000 000 000 742 094| = 0.000 000 000 742 094


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: 0.000 000 000 742 094.

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) 0.000 000 000 742 094 × 2 = 0 + 0.000 000 001 484 188;
  • 2) 0.000 000 001 484 188 × 2 = 0 + 0.000 000 002 968 376;
  • 3) 0.000 000 002 968 376 × 2 = 0 + 0.000 000 005 936 752;
  • 4) 0.000 000 005 936 752 × 2 = 0 + 0.000 000 011 873 504;
  • 5) 0.000 000 011 873 504 × 2 = 0 + 0.000 000 023 747 008;
  • 6) 0.000 000 023 747 008 × 2 = 0 + 0.000 000 047 494 016;
  • 7) 0.000 000 047 494 016 × 2 = 0 + 0.000 000 094 988 032;
  • 8) 0.000 000 094 988 032 × 2 = 0 + 0.000 000 189 976 064;
  • 9) 0.000 000 189 976 064 × 2 = 0 + 0.000 000 379 952 128;
  • 10) 0.000 000 379 952 128 × 2 = 0 + 0.000 000 759 904 256;
  • 11) 0.000 000 759 904 256 × 2 = 0 + 0.000 001 519 808 512;
  • 12) 0.000 001 519 808 512 × 2 = 0 + 0.000 003 039 617 024;
  • 13) 0.000 003 039 617 024 × 2 = 0 + 0.000 006 079 234 048;
  • 14) 0.000 006 079 234 048 × 2 = 0 + 0.000 012 158 468 096;
  • 15) 0.000 012 158 468 096 × 2 = 0 + 0.000 024 316 936 192;
  • 16) 0.000 024 316 936 192 × 2 = 0 + 0.000 048 633 872 384;
  • 17) 0.000 048 633 872 384 × 2 = 0 + 0.000 097 267 744 768;
  • 18) 0.000 097 267 744 768 × 2 = 0 + 0.000 194 535 489 536;
  • 19) 0.000 194 535 489 536 × 2 = 0 + 0.000 389 070 979 072;
  • 20) 0.000 389 070 979 072 × 2 = 0 + 0.000 778 141 958 144;
  • 21) 0.000 778 141 958 144 × 2 = 0 + 0.001 556 283 916 288;
  • 22) 0.001 556 283 916 288 × 2 = 0 + 0.003 112 567 832 576;
  • 23) 0.003 112 567 832 576 × 2 = 0 + 0.006 225 135 665 152;
  • 24) 0.006 225 135 665 152 × 2 = 0 + 0.012 450 271 330 304;
  • 25) 0.012 450 271 330 304 × 2 = 0 + 0.024 900 542 660 608;
  • 26) 0.024 900 542 660 608 × 2 = 0 + 0.049 801 085 321 216;
  • 27) 0.049 801 085 321 216 × 2 = 0 + 0.099 602 170 642 432;
  • 28) 0.099 602 170 642 432 × 2 = 0 + 0.199 204 341 284 864;
  • 29) 0.199 204 341 284 864 × 2 = 0 + 0.398 408 682 569 728;
  • 30) 0.398 408 682 569 728 × 2 = 0 + 0.796 817 365 139 456;
  • 31) 0.796 817 365 139 456 × 2 = 1 + 0.593 634 730 278 912;
  • 32) 0.593 634 730 278 912 × 2 = 1 + 0.187 269 460 557 824;
  • 33) 0.187 269 460 557 824 × 2 = 0 + 0.374 538 921 115 648;
  • 34) 0.374 538 921 115 648 × 2 = 0 + 0.749 077 842 231 296;
  • 35) 0.749 077 842 231 296 × 2 = 1 + 0.498 155 684 462 592;
  • 36) 0.498 155 684 462 592 × 2 = 0 + 0.996 311 368 925 184;
  • 37) 0.996 311 368 925 184 × 2 = 1 + 0.992 622 737 850 368;
  • 38) 0.992 622 737 850 368 × 2 = 1 + 0.985 245 475 700 736;
  • 39) 0.985 245 475 700 736 × 2 = 1 + 0.970 490 951 401 472;
  • 40) 0.970 490 951 401 472 × 2 = 1 + 0.940 981 902 802 944;
  • 41) 0.940 981 902 802 944 × 2 = 1 + 0.881 963 805 605 888;
  • 42) 0.881 963 805 605 888 × 2 = 1 + 0.763 927 611 211 776;
  • 43) 0.763 927 611 211 776 × 2 = 1 + 0.527 855 222 423 552;
  • 44) 0.527 855 222 423 552 × 2 = 1 + 0.055 710 444 847 104;
  • 45) 0.055 710 444 847 104 × 2 = 0 + 0.111 420 889 694 208;
  • 46) 0.111 420 889 694 208 × 2 = 0 + 0.222 841 779 388 416;
  • 47) 0.222 841 779 388 416 × 2 = 0 + 0.445 683 558 776 832;
  • 48) 0.445 683 558 776 832 × 2 = 0 + 0.891 367 117 553 664;
  • 49) 0.891 367 117 553 664 × 2 = 1 + 0.782 734 235 107 328;
  • 50) 0.782 734 235 107 328 × 2 = 1 + 0.565 468 470 214 656;
  • 51) 0.565 468 470 214 656 × 2 = 1 + 0.130 936 940 429 312;
  • 52) 0.130 936 940 429 312 × 2 = 0 + 0.261 873 880 858 624;
  • 53) 0.261 873 880 858 624 × 2 = 0 + 0.523 747 761 717 248;
  • 54) 0.523 747 761 717 248 × 2 = 1 + 0.047 495 523 434 496;

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:


0.000 000 000 742 094(10) =

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

6. Positive number before normalization:

0.000 000 000 742 094(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 0000 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:


0.000 000 000 742 094(10) =

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

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

1.1001 0111 1111 1000 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 1111 1000 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 1100 0011 1001 =

100 1011 1111 1100 0011 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 1100 0011 1001


Decimal number -0.000 000 000 742 094 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1100 0011 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