-0.000 000 000 742 147 676 043 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 043(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
-0.000 000 000 742 147 676 043(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:

|-0.000 000 000 742 147 676 043| = 0.000 000 000 742 147 676 043


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: 0.000 000 000 742 147 676 043.

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) 0.000 000 000 742 147 676 043 × 2 = 0 + 0.000 000 001 484 295 352 086;
  • 2) 0.000 000 001 484 295 352 086 × 2 = 0 + 0.000 000 002 968 590 704 172;
  • 3) 0.000 000 002 968 590 704 172 × 2 = 0 + 0.000 000 005 937 181 408 344;
  • 4) 0.000 000 005 937 181 408 344 × 2 = 0 + 0.000 000 011 874 362 816 688;
  • 5) 0.000 000 011 874 362 816 688 × 2 = 0 + 0.000 000 023 748 725 633 376;
  • 6) 0.000 000 023 748 725 633 376 × 2 = 0 + 0.000 000 047 497 451 266 752;
  • 7) 0.000 000 047 497 451 266 752 × 2 = 0 + 0.000 000 094 994 902 533 504;
  • 8) 0.000 000 094 994 902 533 504 × 2 = 0 + 0.000 000 189 989 805 067 008;
  • 9) 0.000 000 189 989 805 067 008 × 2 = 0 + 0.000 000 379 979 610 134 016;
  • 10) 0.000 000 379 979 610 134 016 × 2 = 0 + 0.000 000 759 959 220 268 032;
  • 11) 0.000 000 759 959 220 268 032 × 2 = 0 + 0.000 001 519 918 440 536 064;
  • 12) 0.000 001 519 918 440 536 064 × 2 = 0 + 0.000 003 039 836 881 072 128;
  • 13) 0.000 003 039 836 881 072 128 × 2 = 0 + 0.000 006 079 673 762 144 256;
  • 14) 0.000 006 079 673 762 144 256 × 2 = 0 + 0.000 012 159 347 524 288 512;
  • 15) 0.000 012 159 347 524 288 512 × 2 = 0 + 0.000 024 318 695 048 577 024;
  • 16) 0.000 024 318 695 048 577 024 × 2 = 0 + 0.000 048 637 390 097 154 048;
  • 17) 0.000 048 637 390 097 154 048 × 2 = 0 + 0.000 097 274 780 194 308 096;
  • 18) 0.000 097 274 780 194 308 096 × 2 = 0 + 0.000 194 549 560 388 616 192;
  • 19) 0.000 194 549 560 388 616 192 × 2 = 0 + 0.000 389 099 120 777 232 384;
  • 20) 0.000 389 099 120 777 232 384 × 2 = 0 + 0.000 778 198 241 554 464 768;
  • 21) 0.000 778 198 241 554 464 768 × 2 = 0 + 0.001 556 396 483 108 929 536;
  • 22) 0.001 556 396 483 108 929 536 × 2 = 0 + 0.003 112 792 966 217 859 072;
  • 23) 0.003 112 792 966 217 859 072 × 2 = 0 + 0.006 225 585 932 435 718 144;
  • 24) 0.006 225 585 932 435 718 144 × 2 = 0 + 0.012 451 171 864 871 436 288;
  • 25) 0.012 451 171 864 871 436 288 × 2 = 0 + 0.024 902 343 729 742 872 576;
  • 26) 0.024 902 343 729 742 872 576 × 2 = 0 + 0.049 804 687 459 485 745 152;
  • 27) 0.049 804 687 459 485 745 152 × 2 = 0 + 0.099 609 374 918 971 490 304;
  • 28) 0.099 609 374 918 971 490 304 × 2 = 0 + 0.199 218 749 837 942 980 608;
  • 29) 0.199 218 749 837 942 980 608 × 2 = 0 + 0.398 437 499 675 885 961 216;
  • 30) 0.398 437 499 675 885 961 216 × 2 = 0 + 0.796 874 999 351 771 922 432;
  • 31) 0.796 874 999 351 771 922 432 × 2 = 1 + 0.593 749 998 703 543 844 864;
  • 32) 0.593 749 998 703 543 844 864 × 2 = 1 + 0.187 499 997 407 087 689 728;
  • 33) 0.187 499 997 407 087 689 728 × 2 = 0 + 0.374 999 994 814 175 379 456;
  • 34) 0.374 999 994 814 175 379 456 × 2 = 0 + 0.749 999 989 628 350 758 912;
  • 35) 0.749 999 989 628 350 758 912 × 2 = 1 + 0.499 999 979 256 701 517 824;
  • 36) 0.499 999 979 256 701 517 824 × 2 = 0 + 0.999 999 958 513 403 035 648;
  • 37) 0.999 999 958 513 403 035 648 × 2 = 1 + 0.999 999 917 026 806 071 296;
  • 38) 0.999 999 917 026 806 071 296 × 2 = 1 + 0.999 999 834 053 612 142 592;
  • 39) 0.999 999 834 053 612 142 592 × 2 = 1 + 0.999 999 668 107 224 285 184;
  • 40) 0.999 999 668 107 224 285 184 × 2 = 1 + 0.999 999 336 214 448 570 368;
  • 41) 0.999 999 336 214 448 570 368 × 2 = 1 + 0.999 998 672 428 897 140 736;
  • 42) 0.999 998 672 428 897 140 736 × 2 = 1 + 0.999 997 344 857 794 281 472;
  • 43) 0.999 997 344 857 794 281 472 × 2 = 1 + 0.999 994 689 715 588 562 944;
  • 44) 0.999 994 689 715 588 562 944 × 2 = 1 + 0.999 989 379 431 177 125 888;
  • 45) 0.999 989 379 431 177 125 888 × 2 = 1 + 0.999 978 758 862 354 251 776;
  • 46) 0.999 978 758 862 354 251 776 × 2 = 1 + 0.999 957 517 724 708 503 552;
  • 47) 0.999 957 517 724 708 503 552 × 2 = 1 + 0.999 915 035 449 417 007 104;
  • 48) 0.999 915 035 449 417 007 104 × 2 = 1 + 0.999 830 070 898 834 014 208;
  • 49) 0.999 830 070 898 834 014 208 × 2 = 1 + 0.999 660 141 797 668 028 416;
  • 50) 0.999 660 141 797 668 028 416 × 2 = 1 + 0.999 320 283 595 336 056 832;
  • 51) 0.999 320 283 595 336 056 832 × 2 = 1 + 0.998 640 567 190 672 113 664;
  • 52) 0.998 640 567 190 672 113 664 × 2 = 1 + 0.997 281 134 381 344 227 328;
  • 53) 0.997 281 134 381 344 227 328 × 2 = 1 + 0.994 562 268 762 688 454 656;
  • 54) 0.994 562 268 762 688 454 656 × 2 = 1 + 0.989 124 537 525 376 909 312;

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:


0.000 000 000 742 147 676 043(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 043(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:


0.000 000 000 742 147 676 043(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 -0.000 000 000 742 147 676 043 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 107 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