-0.000 000 000 742 147 428 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 428(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
-0.000 000 000 742 147 428(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:

|-0.000 000 000 742 147 428| = 0.000 000 000 742 147 428


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: 0.000 000 000 742 147 428.

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) 0.000 000 000 742 147 428 × 2 = 0 + 0.000 000 001 484 294 856;
  • 2) 0.000 000 001 484 294 856 × 2 = 0 + 0.000 000 002 968 589 712;
  • 3) 0.000 000 002 968 589 712 × 2 = 0 + 0.000 000 005 937 179 424;
  • 4) 0.000 000 005 937 179 424 × 2 = 0 + 0.000 000 011 874 358 848;
  • 5) 0.000 000 011 874 358 848 × 2 = 0 + 0.000 000 023 748 717 696;
  • 6) 0.000 000 023 748 717 696 × 2 = 0 + 0.000 000 047 497 435 392;
  • 7) 0.000 000 047 497 435 392 × 2 = 0 + 0.000 000 094 994 870 784;
  • 8) 0.000 000 094 994 870 784 × 2 = 0 + 0.000 000 189 989 741 568;
  • 9) 0.000 000 189 989 741 568 × 2 = 0 + 0.000 000 379 979 483 136;
  • 10) 0.000 000 379 979 483 136 × 2 = 0 + 0.000 000 759 958 966 272;
  • 11) 0.000 000 759 958 966 272 × 2 = 0 + 0.000 001 519 917 932 544;
  • 12) 0.000 001 519 917 932 544 × 2 = 0 + 0.000 003 039 835 865 088;
  • 13) 0.000 003 039 835 865 088 × 2 = 0 + 0.000 006 079 671 730 176;
  • 14) 0.000 006 079 671 730 176 × 2 = 0 + 0.000 012 159 343 460 352;
  • 15) 0.000 012 159 343 460 352 × 2 = 0 + 0.000 024 318 686 920 704;
  • 16) 0.000 024 318 686 920 704 × 2 = 0 + 0.000 048 637 373 841 408;
  • 17) 0.000 048 637 373 841 408 × 2 = 0 + 0.000 097 274 747 682 816;
  • 18) 0.000 097 274 747 682 816 × 2 = 0 + 0.000 194 549 495 365 632;
  • 19) 0.000 194 549 495 365 632 × 2 = 0 + 0.000 389 098 990 731 264;
  • 20) 0.000 389 098 990 731 264 × 2 = 0 + 0.000 778 197 981 462 528;
  • 21) 0.000 778 197 981 462 528 × 2 = 0 + 0.001 556 395 962 925 056;
  • 22) 0.001 556 395 962 925 056 × 2 = 0 + 0.003 112 791 925 850 112;
  • 23) 0.003 112 791 925 850 112 × 2 = 0 + 0.006 225 583 851 700 224;
  • 24) 0.006 225 583 851 700 224 × 2 = 0 + 0.012 451 167 703 400 448;
  • 25) 0.012 451 167 703 400 448 × 2 = 0 + 0.024 902 335 406 800 896;
  • 26) 0.024 902 335 406 800 896 × 2 = 0 + 0.049 804 670 813 601 792;
  • 27) 0.049 804 670 813 601 792 × 2 = 0 + 0.099 609 341 627 203 584;
  • 28) 0.099 609 341 627 203 584 × 2 = 0 + 0.199 218 683 254 407 168;
  • 29) 0.199 218 683 254 407 168 × 2 = 0 + 0.398 437 366 508 814 336;
  • 30) 0.398 437 366 508 814 336 × 2 = 0 + 0.796 874 733 017 628 672;
  • 31) 0.796 874 733 017 628 672 × 2 = 1 + 0.593 749 466 035 257 344;
  • 32) 0.593 749 466 035 257 344 × 2 = 1 + 0.187 498 932 070 514 688;
  • 33) 0.187 498 932 070 514 688 × 2 = 0 + 0.374 997 864 141 029 376;
  • 34) 0.374 997 864 141 029 376 × 2 = 0 + 0.749 995 728 282 058 752;
  • 35) 0.749 995 728 282 058 752 × 2 = 1 + 0.499 991 456 564 117 504;
  • 36) 0.499 991 456 564 117 504 × 2 = 0 + 0.999 982 913 128 235 008;
  • 37) 0.999 982 913 128 235 008 × 2 = 1 + 0.999 965 826 256 470 016;
  • 38) 0.999 965 826 256 470 016 × 2 = 1 + 0.999 931 652 512 940 032;
  • 39) 0.999 931 652 512 940 032 × 2 = 1 + 0.999 863 305 025 880 064;
  • 40) 0.999 863 305 025 880 064 × 2 = 1 + 0.999 726 610 051 760 128;
  • 41) 0.999 726 610 051 760 128 × 2 = 1 + 0.999 453 220 103 520 256;
  • 42) 0.999 453 220 103 520 256 × 2 = 1 + 0.998 906 440 207 040 512;
  • 43) 0.998 906 440 207 040 512 × 2 = 1 + 0.997 812 880 414 081 024;
  • 44) 0.997 812 880 414 081 024 × 2 = 1 + 0.995 625 760 828 162 048;
  • 45) 0.995 625 760 828 162 048 × 2 = 1 + 0.991 251 521 656 324 096;
  • 46) 0.991 251 521 656 324 096 × 2 = 1 + 0.982 503 043 312 648 192;
  • 47) 0.982 503 043 312 648 192 × 2 = 1 + 0.965 006 086 625 296 384;
  • 48) 0.965 006 086 625 296 384 × 2 = 1 + 0.930 012 173 250 592 768;
  • 49) 0.930 012 173 250 592 768 × 2 = 1 + 0.860 024 346 501 185 536;
  • 50) 0.860 024 346 501 185 536 × 2 = 1 + 0.720 048 693 002 371 072;
  • 51) 0.720 048 693 002 371 072 × 2 = 1 + 0.440 097 386 004 742 144;
  • 52) 0.440 097 386 004 742 144 × 2 = 0 + 0.880 194 772 009 484 288;
  • 53) 0.880 194 772 009 484 288 × 2 = 1 + 0.760 389 544 018 968 576;
  • 54) 0.760 389 544 018 968 576 × 2 = 1 + 0.520 779 088 037 937 152;

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:


0.000 000 000 742 147 428(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 428(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 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:


0.000 000 000 742 147 428(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2) × 20 =

1.1001 0111 1111 1111 1111 011(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 1111 1111 011


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 1111 1111 1011 =

100 1011 1111 1111 1111 1011


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 1111 1111 1011


Decimal number -0.000 000 000 742 147 428 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1011

Share

» Convert decimal -0.000 000 000 742 147 454 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

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