-0.000 000 000 742 147 676 744 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 744(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
-0.000 000 000 742 147 676 744(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:

|-0.000 000 000 742 147 676 744| = 0.000 000 000 742 147 676 744


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: 0.000 000 000 742 147 676 744.

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) 0.000 000 000 742 147 676 744 × 2 = 0 + 0.000 000 001 484 295 353 488;
  • 2) 0.000 000 001 484 295 353 488 × 2 = 0 + 0.000 000 002 968 590 706 976;
  • 3) 0.000 000 002 968 590 706 976 × 2 = 0 + 0.000 000 005 937 181 413 952;
  • 4) 0.000 000 005 937 181 413 952 × 2 = 0 + 0.000 000 011 874 362 827 904;
  • 5) 0.000 000 011 874 362 827 904 × 2 = 0 + 0.000 000 023 748 725 655 808;
  • 6) 0.000 000 023 748 725 655 808 × 2 = 0 + 0.000 000 047 497 451 311 616;
  • 7) 0.000 000 047 497 451 311 616 × 2 = 0 + 0.000 000 094 994 902 623 232;
  • 8) 0.000 000 094 994 902 623 232 × 2 = 0 + 0.000 000 189 989 805 246 464;
  • 9) 0.000 000 189 989 805 246 464 × 2 = 0 + 0.000 000 379 979 610 492 928;
  • 10) 0.000 000 379 979 610 492 928 × 2 = 0 + 0.000 000 759 959 220 985 856;
  • 11) 0.000 000 759 959 220 985 856 × 2 = 0 + 0.000 001 519 918 441 971 712;
  • 12) 0.000 001 519 918 441 971 712 × 2 = 0 + 0.000 003 039 836 883 943 424;
  • 13) 0.000 003 039 836 883 943 424 × 2 = 0 + 0.000 006 079 673 767 886 848;
  • 14) 0.000 006 079 673 767 886 848 × 2 = 0 + 0.000 012 159 347 535 773 696;
  • 15) 0.000 012 159 347 535 773 696 × 2 = 0 + 0.000 024 318 695 071 547 392;
  • 16) 0.000 024 318 695 071 547 392 × 2 = 0 + 0.000 048 637 390 143 094 784;
  • 17) 0.000 048 637 390 143 094 784 × 2 = 0 + 0.000 097 274 780 286 189 568;
  • 18) 0.000 097 274 780 286 189 568 × 2 = 0 + 0.000 194 549 560 572 379 136;
  • 19) 0.000 194 549 560 572 379 136 × 2 = 0 + 0.000 389 099 121 144 758 272;
  • 20) 0.000 389 099 121 144 758 272 × 2 = 0 + 0.000 778 198 242 289 516 544;
  • 21) 0.000 778 198 242 289 516 544 × 2 = 0 + 0.001 556 396 484 579 033 088;
  • 22) 0.001 556 396 484 579 033 088 × 2 = 0 + 0.003 112 792 969 158 066 176;
  • 23) 0.003 112 792 969 158 066 176 × 2 = 0 + 0.006 225 585 938 316 132 352;
  • 24) 0.006 225 585 938 316 132 352 × 2 = 0 + 0.012 451 171 876 632 264 704;
  • 25) 0.012 451 171 876 632 264 704 × 2 = 0 + 0.024 902 343 753 264 529 408;
  • 26) 0.024 902 343 753 264 529 408 × 2 = 0 + 0.049 804 687 506 529 058 816;
  • 27) 0.049 804 687 506 529 058 816 × 2 = 0 + 0.099 609 375 013 058 117 632;
  • 28) 0.099 609 375 013 058 117 632 × 2 = 0 + 0.199 218 750 026 116 235 264;
  • 29) 0.199 218 750 026 116 235 264 × 2 = 0 + 0.398 437 500 052 232 470 528;
  • 30) 0.398 437 500 052 232 470 528 × 2 = 0 + 0.796 875 000 104 464 941 056;
  • 31) 0.796 875 000 104 464 941 056 × 2 = 1 + 0.593 750 000 208 929 882 112;
  • 32) 0.593 750 000 208 929 882 112 × 2 = 1 + 0.187 500 000 417 859 764 224;
  • 33) 0.187 500 000 417 859 764 224 × 2 = 0 + 0.375 000 000 835 719 528 448;
  • 34) 0.375 000 000 835 719 528 448 × 2 = 0 + 0.750 000 001 671 439 056 896;
  • 35) 0.750 000 001 671 439 056 896 × 2 = 1 + 0.500 000 003 342 878 113 792;
  • 36) 0.500 000 003 342 878 113 792 × 2 = 1 + 0.000 000 006 685 756 227 584;
  • 37) 0.000 000 006 685 756 227 584 × 2 = 0 + 0.000 000 013 371 512 455 168;
  • 38) 0.000 000 013 371 512 455 168 × 2 = 0 + 0.000 000 026 743 024 910 336;
  • 39) 0.000 000 026 743 024 910 336 × 2 = 0 + 0.000 000 053 486 049 820 672;
  • 40) 0.000 000 053 486 049 820 672 × 2 = 0 + 0.000 000 106 972 099 641 344;
  • 41) 0.000 000 106 972 099 641 344 × 2 = 0 + 0.000 000 213 944 199 282 688;
  • 42) 0.000 000 213 944 199 282 688 × 2 = 0 + 0.000 000 427 888 398 565 376;
  • 43) 0.000 000 427 888 398 565 376 × 2 = 0 + 0.000 000 855 776 797 130 752;
  • 44) 0.000 000 855 776 797 130 752 × 2 = 0 + 0.000 001 711 553 594 261 504;
  • 45) 0.000 001 711 553 594 261 504 × 2 = 0 + 0.000 003 423 107 188 523 008;
  • 46) 0.000 003 423 107 188 523 008 × 2 = 0 + 0.000 006 846 214 377 046 016;
  • 47) 0.000 006 846 214 377 046 016 × 2 = 0 + 0.000 013 692 428 754 092 032;
  • 48) 0.000 013 692 428 754 092 032 × 2 = 0 + 0.000 027 384 857 508 184 064;
  • 49) 0.000 027 384 857 508 184 064 × 2 = 0 + 0.000 054 769 715 016 368 128;
  • 50) 0.000 054 769 715 016 368 128 × 2 = 0 + 0.000 109 539 430 032 736 256;
  • 51) 0.000 109 539 430 032 736 256 × 2 = 0 + 0.000 219 078 860 065 472 512;
  • 52) 0.000 219 078 860 065 472 512 × 2 = 0 + 0.000 438 157 720 130 945 024;
  • 53) 0.000 438 157 720 130 945 024 × 2 = 0 + 0.000 876 315 440 261 890 048;
  • 54) 0.000 876 315 440 261 890 048 × 2 = 0 + 0.001 752 630 880 523 780 096;

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:


0.000 000 000 742 147 676 744(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 744(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 744(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 744 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

Share

» Convert decimal -0.000 000 000 742 147 676 814 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: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 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