-0.000 000 000 742 117 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 117(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
-0.000 000 000 742 117(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:

|-0.000 000 000 742 117| = 0.000 000 000 742 117


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: 0.000 000 000 742 117.

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) 0.000 000 000 742 117 × 2 = 0 + 0.000 000 001 484 234;
  • 2) 0.000 000 001 484 234 × 2 = 0 + 0.000 000 002 968 468;
  • 3) 0.000 000 002 968 468 × 2 = 0 + 0.000 000 005 936 936;
  • 4) 0.000 000 005 936 936 × 2 = 0 + 0.000 000 011 873 872;
  • 5) 0.000 000 011 873 872 × 2 = 0 + 0.000 000 023 747 744;
  • 6) 0.000 000 023 747 744 × 2 = 0 + 0.000 000 047 495 488;
  • 7) 0.000 000 047 495 488 × 2 = 0 + 0.000 000 094 990 976;
  • 8) 0.000 000 094 990 976 × 2 = 0 + 0.000 000 189 981 952;
  • 9) 0.000 000 189 981 952 × 2 = 0 + 0.000 000 379 963 904;
  • 10) 0.000 000 379 963 904 × 2 = 0 + 0.000 000 759 927 808;
  • 11) 0.000 000 759 927 808 × 2 = 0 + 0.000 001 519 855 616;
  • 12) 0.000 001 519 855 616 × 2 = 0 + 0.000 003 039 711 232;
  • 13) 0.000 003 039 711 232 × 2 = 0 + 0.000 006 079 422 464;
  • 14) 0.000 006 079 422 464 × 2 = 0 + 0.000 012 158 844 928;
  • 15) 0.000 012 158 844 928 × 2 = 0 + 0.000 024 317 689 856;
  • 16) 0.000 024 317 689 856 × 2 = 0 + 0.000 048 635 379 712;
  • 17) 0.000 048 635 379 712 × 2 = 0 + 0.000 097 270 759 424;
  • 18) 0.000 097 270 759 424 × 2 = 0 + 0.000 194 541 518 848;
  • 19) 0.000 194 541 518 848 × 2 = 0 + 0.000 389 083 037 696;
  • 20) 0.000 389 083 037 696 × 2 = 0 + 0.000 778 166 075 392;
  • 21) 0.000 778 166 075 392 × 2 = 0 + 0.001 556 332 150 784;
  • 22) 0.001 556 332 150 784 × 2 = 0 + 0.003 112 664 301 568;
  • 23) 0.003 112 664 301 568 × 2 = 0 + 0.006 225 328 603 136;
  • 24) 0.006 225 328 603 136 × 2 = 0 + 0.012 450 657 206 272;
  • 25) 0.012 450 657 206 272 × 2 = 0 + 0.024 901 314 412 544;
  • 26) 0.024 901 314 412 544 × 2 = 0 + 0.049 802 628 825 088;
  • 27) 0.049 802 628 825 088 × 2 = 0 + 0.099 605 257 650 176;
  • 28) 0.099 605 257 650 176 × 2 = 0 + 0.199 210 515 300 352;
  • 29) 0.199 210 515 300 352 × 2 = 0 + 0.398 421 030 600 704;
  • 30) 0.398 421 030 600 704 × 2 = 0 + 0.796 842 061 201 408;
  • 31) 0.796 842 061 201 408 × 2 = 1 + 0.593 684 122 402 816;
  • 32) 0.593 684 122 402 816 × 2 = 1 + 0.187 368 244 805 632;
  • 33) 0.187 368 244 805 632 × 2 = 0 + 0.374 736 489 611 264;
  • 34) 0.374 736 489 611 264 × 2 = 0 + 0.749 472 979 222 528;
  • 35) 0.749 472 979 222 528 × 2 = 1 + 0.498 945 958 445 056;
  • 36) 0.498 945 958 445 056 × 2 = 0 + 0.997 891 916 890 112;
  • 37) 0.997 891 916 890 112 × 2 = 1 + 0.995 783 833 780 224;
  • 38) 0.995 783 833 780 224 × 2 = 1 + 0.991 567 667 560 448;
  • 39) 0.991 567 667 560 448 × 2 = 1 + 0.983 135 335 120 896;
  • 40) 0.983 135 335 120 896 × 2 = 1 + 0.966 270 670 241 792;
  • 41) 0.966 270 670 241 792 × 2 = 1 + 0.932 541 340 483 584;
  • 42) 0.932 541 340 483 584 × 2 = 1 + 0.865 082 680 967 168;
  • 43) 0.865 082 680 967 168 × 2 = 1 + 0.730 165 361 934 336;
  • 44) 0.730 165 361 934 336 × 2 = 1 + 0.460 330 723 868 672;
  • 45) 0.460 330 723 868 672 × 2 = 0 + 0.920 661 447 737 344;
  • 46) 0.920 661 447 737 344 × 2 = 1 + 0.841 322 895 474 688;
  • 47) 0.841 322 895 474 688 × 2 = 1 + 0.682 645 790 949 376;
  • 48) 0.682 645 790 949 376 × 2 = 1 + 0.365 291 581 898 752;
  • 49) 0.365 291 581 898 752 × 2 = 0 + 0.730 583 163 797 504;
  • 50) 0.730 583 163 797 504 × 2 = 1 + 0.461 166 327 595 008;
  • 51) 0.461 166 327 595 008 × 2 = 0 + 0.922 332 655 190 016;
  • 52) 0.922 332 655 190 016 × 2 = 1 + 0.844 665 310 380 032;
  • 53) 0.844 665 310 380 032 × 2 = 1 + 0.689 330 620 760 064;
  • 54) 0.689 330 620 760 064 × 2 = 1 + 0.378 661 241 520 128;

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:


0.000 000 000 742 117(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 0111 0101 11(2)

6. Positive number before normalization:

0.000 000 000 742 117(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 0111 0101 11(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:


0.000 000 000 742 117(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 0111 0101 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 0111 0101 11(2) × 20 =

1.1001 0111 1111 1011 1010 111(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 1011 1010 111


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 1101 1101 0111 =

100 1011 1111 1101 1101 0111


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 1101 1101 0111


Decimal number -0.000 000 000 742 117 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1101 1101 0111

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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