-0.000 000 000 742 147 676 646 693 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 693(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
-0.000 000 000 742 147 676 646 693(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:

|-0.000 000 000 742 147 676 646 693| = 0.000 000 000 742 147 676 646 693


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: 0.000 000 000 742 147 676 646 693.

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) 0.000 000 000 742 147 676 646 693 × 2 = 0 + 0.000 000 001 484 295 353 293 386;
  • 2) 0.000 000 001 484 295 353 293 386 × 2 = 0 + 0.000 000 002 968 590 706 586 772;
  • 3) 0.000 000 002 968 590 706 586 772 × 2 = 0 + 0.000 000 005 937 181 413 173 544;
  • 4) 0.000 000 005 937 181 413 173 544 × 2 = 0 + 0.000 000 011 874 362 826 347 088;
  • 5) 0.000 000 011 874 362 826 347 088 × 2 = 0 + 0.000 000 023 748 725 652 694 176;
  • 6) 0.000 000 023 748 725 652 694 176 × 2 = 0 + 0.000 000 047 497 451 305 388 352;
  • 7) 0.000 000 047 497 451 305 388 352 × 2 = 0 + 0.000 000 094 994 902 610 776 704;
  • 8) 0.000 000 094 994 902 610 776 704 × 2 = 0 + 0.000 000 189 989 805 221 553 408;
  • 9) 0.000 000 189 989 805 221 553 408 × 2 = 0 + 0.000 000 379 979 610 443 106 816;
  • 10) 0.000 000 379 979 610 443 106 816 × 2 = 0 + 0.000 000 759 959 220 886 213 632;
  • 11) 0.000 000 759 959 220 886 213 632 × 2 = 0 + 0.000 001 519 918 441 772 427 264;
  • 12) 0.000 001 519 918 441 772 427 264 × 2 = 0 + 0.000 003 039 836 883 544 854 528;
  • 13) 0.000 003 039 836 883 544 854 528 × 2 = 0 + 0.000 006 079 673 767 089 709 056;
  • 14) 0.000 006 079 673 767 089 709 056 × 2 = 0 + 0.000 012 159 347 534 179 418 112;
  • 15) 0.000 012 159 347 534 179 418 112 × 2 = 0 + 0.000 024 318 695 068 358 836 224;
  • 16) 0.000 024 318 695 068 358 836 224 × 2 = 0 + 0.000 048 637 390 136 717 672 448;
  • 17) 0.000 048 637 390 136 717 672 448 × 2 = 0 + 0.000 097 274 780 273 435 344 896;
  • 18) 0.000 097 274 780 273 435 344 896 × 2 = 0 + 0.000 194 549 560 546 870 689 792;
  • 19) 0.000 194 549 560 546 870 689 792 × 2 = 0 + 0.000 389 099 121 093 741 379 584;
  • 20) 0.000 389 099 121 093 741 379 584 × 2 = 0 + 0.000 778 198 242 187 482 759 168;
  • 21) 0.000 778 198 242 187 482 759 168 × 2 = 0 + 0.001 556 396 484 374 965 518 336;
  • 22) 0.001 556 396 484 374 965 518 336 × 2 = 0 + 0.003 112 792 968 749 931 036 672;
  • 23) 0.003 112 792 968 749 931 036 672 × 2 = 0 + 0.006 225 585 937 499 862 073 344;
  • 24) 0.006 225 585 937 499 862 073 344 × 2 = 0 + 0.012 451 171 874 999 724 146 688;
  • 25) 0.012 451 171 874 999 724 146 688 × 2 = 0 + 0.024 902 343 749 999 448 293 376;
  • 26) 0.024 902 343 749 999 448 293 376 × 2 = 0 + 0.049 804 687 499 998 896 586 752;
  • 27) 0.049 804 687 499 998 896 586 752 × 2 = 0 + 0.099 609 374 999 997 793 173 504;
  • 28) 0.099 609 374 999 997 793 173 504 × 2 = 0 + 0.199 218 749 999 995 586 347 008;
  • 29) 0.199 218 749 999 995 586 347 008 × 2 = 0 + 0.398 437 499 999 991 172 694 016;
  • 30) 0.398 437 499 999 991 172 694 016 × 2 = 0 + 0.796 874 999 999 982 345 388 032;
  • 31) 0.796 874 999 999 982 345 388 032 × 2 = 1 + 0.593 749 999 999 964 690 776 064;
  • 32) 0.593 749 999 999 964 690 776 064 × 2 = 1 + 0.187 499 999 999 929 381 552 128;
  • 33) 0.187 499 999 999 929 381 552 128 × 2 = 0 + 0.374 999 999 999 858 763 104 256;
  • 34) 0.374 999 999 999 858 763 104 256 × 2 = 0 + 0.749 999 999 999 717 526 208 512;
  • 35) 0.749 999 999 999 717 526 208 512 × 2 = 1 + 0.499 999 999 999 435 052 417 024;
  • 36) 0.499 999 999 999 435 052 417 024 × 2 = 0 + 0.999 999 999 998 870 104 834 048;
  • 37) 0.999 999 999 998 870 104 834 048 × 2 = 1 + 0.999 999 999 997 740 209 668 096;
  • 38) 0.999 999 999 997 740 209 668 096 × 2 = 1 + 0.999 999 999 995 480 419 336 192;
  • 39) 0.999 999 999 995 480 419 336 192 × 2 = 1 + 0.999 999 999 990 960 838 672 384;
  • 40) 0.999 999 999 990 960 838 672 384 × 2 = 1 + 0.999 999 999 981 921 677 344 768;
  • 41) 0.999 999 999 981 921 677 344 768 × 2 = 1 + 0.999 999 999 963 843 354 689 536;
  • 42) 0.999 999 999 963 843 354 689 536 × 2 = 1 + 0.999 999 999 927 686 709 379 072;
  • 43) 0.999 999 999 927 686 709 379 072 × 2 = 1 + 0.999 999 999 855 373 418 758 144;
  • 44) 0.999 999 999 855 373 418 758 144 × 2 = 1 + 0.999 999 999 710 746 837 516 288;
  • 45) 0.999 999 999 710 746 837 516 288 × 2 = 1 + 0.999 999 999 421 493 675 032 576;
  • 46) 0.999 999 999 421 493 675 032 576 × 2 = 1 + 0.999 999 998 842 987 350 065 152;
  • 47) 0.999 999 998 842 987 350 065 152 × 2 = 1 + 0.999 999 997 685 974 700 130 304;
  • 48) 0.999 999 997 685 974 700 130 304 × 2 = 1 + 0.999 999 995 371 949 400 260 608;
  • 49) 0.999 999 995 371 949 400 260 608 × 2 = 1 + 0.999 999 990 743 898 800 521 216;
  • 50) 0.999 999 990 743 898 800 521 216 × 2 = 1 + 0.999 999 981 487 797 601 042 432;
  • 51) 0.999 999 981 487 797 601 042 432 × 2 = 1 + 0.999 999 962 975 595 202 084 864;
  • 52) 0.999 999 962 975 595 202 084 864 × 2 = 1 + 0.999 999 925 951 190 404 169 728;
  • 53) 0.999 999 925 951 190 404 169 728 × 2 = 1 + 0.999 999 851 902 380 808 339 456;
  • 54) 0.999 999 851 902 380 808 339 456 × 2 = 1 + 0.999 999 703 804 761 616 678 912;

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:


0.000 000 000 742 147 676 646 693(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 693(10) =

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


0.000 000 000 742 147 676 646 693(10) =

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

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

1.1001 0111 1111 1111 1111 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 1111 1111 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 1111 1111 1111 =

100 1011 1111 1111 1111 1111


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 1111


Decimal number -0.000 000 000 742 147 676 646 693 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 646 706 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 Art of Strategic Ambiguity and Concealed Intentions. NotebookLM YouTube Video of 6.59 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