-0.000 000 000 742 147 676 693 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 693(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
-0.000 000 000 742 147 676 693(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:

|-0.000 000 000 742 147 676 693| = 0.000 000 000 742 147 676 693


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: 0.000 000 000 742 147 676 693.

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) 0.000 000 000 742 147 676 693 × 2 = 0 + 0.000 000 001 484 295 353 386;
  • 2) 0.000 000 001 484 295 353 386 × 2 = 0 + 0.000 000 002 968 590 706 772;
  • 3) 0.000 000 002 968 590 706 772 × 2 = 0 + 0.000 000 005 937 181 413 544;
  • 4) 0.000 000 005 937 181 413 544 × 2 = 0 + 0.000 000 011 874 362 827 088;
  • 5) 0.000 000 011 874 362 827 088 × 2 = 0 + 0.000 000 023 748 725 654 176;
  • 6) 0.000 000 023 748 725 654 176 × 2 = 0 + 0.000 000 047 497 451 308 352;
  • 7) 0.000 000 047 497 451 308 352 × 2 = 0 + 0.000 000 094 994 902 616 704;
  • 8) 0.000 000 094 994 902 616 704 × 2 = 0 + 0.000 000 189 989 805 233 408;
  • 9) 0.000 000 189 989 805 233 408 × 2 = 0 + 0.000 000 379 979 610 466 816;
  • 10) 0.000 000 379 979 610 466 816 × 2 = 0 + 0.000 000 759 959 220 933 632;
  • 11) 0.000 000 759 959 220 933 632 × 2 = 0 + 0.000 001 519 918 441 867 264;
  • 12) 0.000 001 519 918 441 867 264 × 2 = 0 + 0.000 003 039 836 883 734 528;
  • 13) 0.000 003 039 836 883 734 528 × 2 = 0 + 0.000 006 079 673 767 469 056;
  • 14) 0.000 006 079 673 767 469 056 × 2 = 0 + 0.000 012 159 347 534 938 112;
  • 15) 0.000 012 159 347 534 938 112 × 2 = 0 + 0.000 024 318 695 069 876 224;
  • 16) 0.000 024 318 695 069 876 224 × 2 = 0 + 0.000 048 637 390 139 752 448;
  • 17) 0.000 048 637 390 139 752 448 × 2 = 0 + 0.000 097 274 780 279 504 896;
  • 18) 0.000 097 274 780 279 504 896 × 2 = 0 + 0.000 194 549 560 559 009 792;
  • 19) 0.000 194 549 560 559 009 792 × 2 = 0 + 0.000 389 099 121 118 019 584;
  • 20) 0.000 389 099 121 118 019 584 × 2 = 0 + 0.000 778 198 242 236 039 168;
  • 21) 0.000 778 198 242 236 039 168 × 2 = 0 + 0.001 556 396 484 472 078 336;
  • 22) 0.001 556 396 484 472 078 336 × 2 = 0 + 0.003 112 792 968 944 156 672;
  • 23) 0.003 112 792 968 944 156 672 × 2 = 0 + 0.006 225 585 937 888 313 344;
  • 24) 0.006 225 585 937 888 313 344 × 2 = 0 + 0.012 451 171 875 776 626 688;
  • 25) 0.012 451 171 875 776 626 688 × 2 = 0 + 0.024 902 343 751 553 253 376;
  • 26) 0.024 902 343 751 553 253 376 × 2 = 0 + 0.049 804 687 503 106 506 752;
  • 27) 0.049 804 687 503 106 506 752 × 2 = 0 + 0.099 609 375 006 213 013 504;
  • 28) 0.099 609 375 006 213 013 504 × 2 = 0 + 0.199 218 750 012 426 027 008;
  • 29) 0.199 218 750 012 426 027 008 × 2 = 0 + 0.398 437 500 024 852 054 016;
  • 30) 0.398 437 500 024 852 054 016 × 2 = 0 + 0.796 875 000 049 704 108 032;
  • 31) 0.796 875 000 049 704 108 032 × 2 = 1 + 0.593 750 000 099 408 216 064;
  • 32) 0.593 750 000 099 408 216 064 × 2 = 1 + 0.187 500 000 198 816 432 128;
  • 33) 0.187 500 000 198 816 432 128 × 2 = 0 + 0.375 000 000 397 632 864 256;
  • 34) 0.375 000 000 397 632 864 256 × 2 = 0 + 0.750 000 000 795 265 728 512;
  • 35) 0.750 000 000 795 265 728 512 × 2 = 1 + 0.500 000 001 590 531 457 024;
  • 36) 0.500 000 001 590 531 457 024 × 2 = 1 + 0.000 000 003 181 062 914 048;
  • 37) 0.000 000 003 181 062 914 048 × 2 = 0 + 0.000 000 006 362 125 828 096;
  • 38) 0.000 000 006 362 125 828 096 × 2 = 0 + 0.000 000 012 724 251 656 192;
  • 39) 0.000 000 012 724 251 656 192 × 2 = 0 + 0.000 000 025 448 503 312 384;
  • 40) 0.000 000 025 448 503 312 384 × 2 = 0 + 0.000 000 050 897 006 624 768;
  • 41) 0.000 000 050 897 006 624 768 × 2 = 0 + 0.000 000 101 794 013 249 536;
  • 42) 0.000 000 101 794 013 249 536 × 2 = 0 + 0.000 000 203 588 026 499 072;
  • 43) 0.000 000 203 588 026 499 072 × 2 = 0 + 0.000 000 407 176 052 998 144;
  • 44) 0.000 000 407 176 052 998 144 × 2 = 0 + 0.000 000 814 352 105 996 288;
  • 45) 0.000 000 814 352 105 996 288 × 2 = 0 + 0.000 001 628 704 211 992 576;
  • 46) 0.000 001 628 704 211 992 576 × 2 = 0 + 0.000 003 257 408 423 985 152;
  • 47) 0.000 003 257 408 423 985 152 × 2 = 0 + 0.000 006 514 816 847 970 304;
  • 48) 0.000 006 514 816 847 970 304 × 2 = 0 + 0.000 013 029 633 695 940 608;
  • 49) 0.000 013 029 633 695 940 608 × 2 = 0 + 0.000 026 059 267 391 881 216;
  • 50) 0.000 026 059 267 391 881 216 × 2 = 0 + 0.000 052 118 534 783 762 432;
  • 51) 0.000 052 118 534 783 762 432 × 2 = 0 + 0.000 104 237 069 567 524 864;
  • 52) 0.000 104 237 069 567 524 864 × 2 = 0 + 0.000 208 474 139 135 049 728;
  • 53) 0.000 208 474 139 135 049 728 × 2 = 0 + 0.000 416 948 278 270 099 456;
  • 54) 0.000 416 948 278 270 099 456 × 2 = 0 + 0.000 833 896 556 540 198 912;

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:


0.000 000 000 742 147 676 693(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 693(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:


0.000 000 000 742 147 676 693(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 -0.000 000 000 742 147 676 693 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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