-0.000 000 000 742 147 676 293 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 293(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
-0.000 000 000 742 147 676 293(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:

|-0.000 000 000 742 147 676 293| = 0.000 000 000 742 147 676 293


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: 0.000 000 000 742 147 676 293.

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) 0.000 000 000 742 147 676 293 × 2 = 0 + 0.000 000 001 484 295 352 586;
  • 2) 0.000 000 001 484 295 352 586 × 2 = 0 + 0.000 000 002 968 590 705 172;
  • 3) 0.000 000 002 968 590 705 172 × 2 = 0 + 0.000 000 005 937 181 410 344;
  • 4) 0.000 000 005 937 181 410 344 × 2 = 0 + 0.000 000 011 874 362 820 688;
  • 5) 0.000 000 011 874 362 820 688 × 2 = 0 + 0.000 000 023 748 725 641 376;
  • 6) 0.000 000 023 748 725 641 376 × 2 = 0 + 0.000 000 047 497 451 282 752;
  • 7) 0.000 000 047 497 451 282 752 × 2 = 0 + 0.000 000 094 994 902 565 504;
  • 8) 0.000 000 094 994 902 565 504 × 2 = 0 + 0.000 000 189 989 805 131 008;
  • 9) 0.000 000 189 989 805 131 008 × 2 = 0 + 0.000 000 379 979 610 262 016;
  • 10) 0.000 000 379 979 610 262 016 × 2 = 0 + 0.000 000 759 959 220 524 032;
  • 11) 0.000 000 759 959 220 524 032 × 2 = 0 + 0.000 001 519 918 441 048 064;
  • 12) 0.000 001 519 918 441 048 064 × 2 = 0 + 0.000 003 039 836 882 096 128;
  • 13) 0.000 003 039 836 882 096 128 × 2 = 0 + 0.000 006 079 673 764 192 256;
  • 14) 0.000 006 079 673 764 192 256 × 2 = 0 + 0.000 012 159 347 528 384 512;
  • 15) 0.000 012 159 347 528 384 512 × 2 = 0 + 0.000 024 318 695 056 769 024;
  • 16) 0.000 024 318 695 056 769 024 × 2 = 0 + 0.000 048 637 390 113 538 048;
  • 17) 0.000 048 637 390 113 538 048 × 2 = 0 + 0.000 097 274 780 227 076 096;
  • 18) 0.000 097 274 780 227 076 096 × 2 = 0 + 0.000 194 549 560 454 152 192;
  • 19) 0.000 194 549 560 454 152 192 × 2 = 0 + 0.000 389 099 120 908 304 384;
  • 20) 0.000 389 099 120 908 304 384 × 2 = 0 + 0.000 778 198 241 816 608 768;
  • 21) 0.000 778 198 241 816 608 768 × 2 = 0 + 0.001 556 396 483 633 217 536;
  • 22) 0.001 556 396 483 633 217 536 × 2 = 0 + 0.003 112 792 967 266 435 072;
  • 23) 0.003 112 792 967 266 435 072 × 2 = 0 + 0.006 225 585 934 532 870 144;
  • 24) 0.006 225 585 934 532 870 144 × 2 = 0 + 0.012 451 171 869 065 740 288;
  • 25) 0.012 451 171 869 065 740 288 × 2 = 0 + 0.024 902 343 738 131 480 576;
  • 26) 0.024 902 343 738 131 480 576 × 2 = 0 + 0.049 804 687 476 262 961 152;
  • 27) 0.049 804 687 476 262 961 152 × 2 = 0 + 0.099 609 374 952 525 922 304;
  • 28) 0.099 609 374 952 525 922 304 × 2 = 0 + 0.199 218 749 905 051 844 608;
  • 29) 0.199 218 749 905 051 844 608 × 2 = 0 + 0.398 437 499 810 103 689 216;
  • 30) 0.398 437 499 810 103 689 216 × 2 = 0 + 0.796 874 999 620 207 378 432;
  • 31) 0.796 874 999 620 207 378 432 × 2 = 1 + 0.593 749 999 240 414 756 864;
  • 32) 0.593 749 999 240 414 756 864 × 2 = 1 + 0.187 499 998 480 829 513 728;
  • 33) 0.187 499 998 480 829 513 728 × 2 = 0 + 0.374 999 996 961 659 027 456;
  • 34) 0.374 999 996 961 659 027 456 × 2 = 0 + 0.749 999 993 923 318 054 912;
  • 35) 0.749 999 993 923 318 054 912 × 2 = 1 + 0.499 999 987 846 636 109 824;
  • 36) 0.499 999 987 846 636 109 824 × 2 = 0 + 0.999 999 975 693 272 219 648;
  • 37) 0.999 999 975 693 272 219 648 × 2 = 1 + 0.999 999 951 386 544 439 296;
  • 38) 0.999 999 951 386 544 439 296 × 2 = 1 + 0.999 999 902 773 088 878 592;
  • 39) 0.999 999 902 773 088 878 592 × 2 = 1 + 0.999 999 805 546 177 757 184;
  • 40) 0.999 999 805 546 177 757 184 × 2 = 1 + 0.999 999 611 092 355 514 368;
  • 41) 0.999 999 611 092 355 514 368 × 2 = 1 + 0.999 999 222 184 711 028 736;
  • 42) 0.999 999 222 184 711 028 736 × 2 = 1 + 0.999 998 444 369 422 057 472;
  • 43) 0.999 998 444 369 422 057 472 × 2 = 1 + 0.999 996 888 738 844 114 944;
  • 44) 0.999 996 888 738 844 114 944 × 2 = 1 + 0.999 993 777 477 688 229 888;
  • 45) 0.999 993 777 477 688 229 888 × 2 = 1 + 0.999 987 554 955 376 459 776;
  • 46) 0.999 987 554 955 376 459 776 × 2 = 1 + 0.999 975 109 910 752 919 552;
  • 47) 0.999 975 109 910 752 919 552 × 2 = 1 + 0.999 950 219 821 505 839 104;
  • 48) 0.999 950 219 821 505 839 104 × 2 = 1 + 0.999 900 439 643 011 678 208;
  • 49) 0.999 900 439 643 011 678 208 × 2 = 1 + 0.999 800 879 286 023 356 416;
  • 50) 0.999 800 879 286 023 356 416 × 2 = 1 + 0.999 601 758 572 046 712 832;
  • 51) 0.999 601 758 572 046 712 832 × 2 = 1 + 0.999 203 517 144 093 425 664;
  • 52) 0.999 203 517 144 093 425 664 × 2 = 1 + 0.998 407 034 288 186 851 328;
  • 53) 0.998 407 034 288 186 851 328 × 2 = 1 + 0.996 814 068 576 373 702 656;
  • 54) 0.996 814 068 576 373 702 656 × 2 = 1 + 0.993 628 137 152 747 405 312;

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:


0.000 000 000 742 147 676 293(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 293(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:


0.000 000 000 742 147 676 293(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 -0.000 000 000 742 147 676 293 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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