-0.000 000 000 742 147 676 451 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 451(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
-0.000 000 000 742 147 676 451(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:

|-0.000 000 000 742 147 676 451| = 0.000 000 000 742 147 676 451


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: 0.000 000 000 742 147 676 451.

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) 0.000 000 000 742 147 676 451 × 2 = 0 + 0.000 000 001 484 295 352 902;
  • 2) 0.000 000 001 484 295 352 902 × 2 = 0 + 0.000 000 002 968 590 705 804;
  • 3) 0.000 000 002 968 590 705 804 × 2 = 0 + 0.000 000 005 937 181 411 608;
  • 4) 0.000 000 005 937 181 411 608 × 2 = 0 + 0.000 000 011 874 362 823 216;
  • 5) 0.000 000 011 874 362 823 216 × 2 = 0 + 0.000 000 023 748 725 646 432;
  • 6) 0.000 000 023 748 725 646 432 × 2 = 0 + 0.000 000 047 497 451 292 864;
  • 7) 0.000 000 047 497 451 292 864 × 2 = 0 + 0.000 000 094 994 902 585 728;
  • 8) 0.000 000 094 994 902 585 728 × 2 = 0 + 0.000 000 189 989 805 171 456;
  • 9) 0.000 000 189 989 805 171 456 × 2 = 0 + 0.000 000 379 979 610 342 912;
  • 10) 0.000 000 379 979 610 342 912 × 2 = 0 + 0.000 000 759 959 220 685 824;
  • 11) 0.000 000 759 959 220 685 824 × 2 = 0 + 0.000 001 519 918 441 371 648;
  • 12) 0.000 001 519 918 441 371 648 × 2 = 0 + 0.000 003 039 836 882 743 296;
  • 13) 0.000 003 039 836 882 743 296 × 2 = 0 + 0.000 006 079 673 765 486 592;
  • 14) 0.000 006 079 673 765 486 592 × 2 = 0 + 0.000 012 159 347 530 973 184;
  • 15) 0.000 012 159 347 530 973 184 × 2 = 0 + 0.000 024 318 695 061 946 368;
  • 16) 0.000 024 318 695 061 946 368 × 2 = 0 + 0.000 048 637 390 123 892 736;
  • 17) 0.000 048 637 390 123 892 736 × 2 = 0 + 0.000 097 274 780 247 785 472;
  • 18) 0.000 097 274 780 247 785 472 × 2 = 0 + 0.000 194 549 560 495 570 944;
  • 19) 0.000 194 549 560 495 570 944 × 2 = 0 + 0.000 389 099 120 991 141 888;
  • 20) 0.000 389 099 120 991 141 888 × 2 = 0 + 0.000 778 198 241 982 283 776;
  • 21) 0.000 778 198 241 982 283 776 × 2 = 0 + 0.001 556 396 483 964 567 552;
  • 22) 0.001 556 396 483 964 567 552 × 2 = 0 + 0.003 112 792 967 929 135 104;
  • 23) 0.003 112 792 967 929 135 104 × 2 = 0 + 0.006 225 585 935 858 270 208;
  • 24) 0.006 225 585 935 858 270 208 × 2 = 0 + 0.012 451 171 871 716 540 416;
  • 25) 0.012 451 171 871 716 540 416 × 2 = 0 + 0.024 902 343 743 433 080 832;
  • 26) 0.024 902 343 743 433 080 832 × 2 = 0 + 0.049 804 687 486 866 161 664;
  • 27) 0.049 804 687 486 866 161 664 × 2 = 0 + 0.099 609 374 973 732 323 328;
  • 28) 0.099 609 374 973 732 323 328 × 2 = 0 + 0.199 218 749 947 464 646 656;
  • 29) 0.199 218 749 947 464 646 656 × 2 = 0 + 0.398 437 499 894 929 293 312;
  • 30) 0.398 437 499 894 929 293 312 × 2 = 0 + 0.796 874 999 789 858 586 624;
  • 31) 0.796 874 999 789 858 586 624 × 2 = 1 + 0.593 749 999 579 717 173 248;
  • 32) 0.593 749 999 579 717 173 248 × 2 = 1 + 0.187 499 999 159 434 346 496;
  • 33) 0.187 499 999 159 434 346 496 × 2 = 0 + 0.374 999 998 318 868 692 992;
  • 34) 0.374 999 998 318 868 692 992 × 2 = 0 + 0.749 999 996 637 737 385 984;
  • 35) 0.749 999 996 637 737 385 984 × 2 = 1 + 0.499 999 993 275 474 771 968;
  • 36) 0.499 999 993 275 474 771 968 × 2 = 0 + 0.999 999 986 550 949 543 936;
  • 37) 0.999 999 986 550 949 543 936 × 2 = 1 + 0.999 999 973 101 899 087 872;
  • 38) 0.999 999 973 101 899 087 872 × 2 = 1 + 0.999 999 946 203 798 175 744;
  • 39) 0.999 999 946 203 798 175 744 × 2 = 1 + 0.999 999 892 407 596 351 488;
  • 40) 0.999 999 892 407 596 351 488 × 2 = 1 + 0.999 999 784 815 192 702 976;
  • 41) 0.999 999 784 815 192 702 976 × 2 = 1 + 0.999 999 569 630 385 405 952;
  • 42) 0.999 999 569 630 385 405 952 × 2 = 1 + 0.999 999 139 260 770 811 904;
  • 43) 0.999 999 139 260 770 811 904 × 2 = 1 + 0.999 998 278 521 541 623 808;
  • 44) 0.999 998 278 521 541 623 808 × 2 = 1 + 0.999 996 557 043 083 247 616;
  • 45) 0.999 996 557 043 083 247 616 × 2 = 1 + 0.999 993 114 086 166 495 232;
  • 46) 0.999 993 114 086 166 495 232 × 2 = 1 + 0.999 986 228 172 332 990 464;
  • 47) 0.999 986 228 172 332 990 464 × 2 = 1 + 0.999 972 456 344 665 980 928;
  • 48) 0.999 972 456 344 665 980 928 × 2 = 1 + 0.999 944 912 689 331 961 856;
  • 49) 0.999 944 912 689 331 961 856 × 2 = 1 + 0.999 889 825 378 663 923 712;
  • 50) 0.999 889 825 378 663 923 712 × 2 = 1 + 0.999 779 650 757 327 847 424;
  • 51) 0.999 779 650 757 327 847 424 × 2 = 1 + 0.999 559 301 514 655 694 848;
  • 52) 0.999 559 301 514 655 694 848 × 2 = 1 + 0.999 118 603 029 311 389 696;
  • 53) 0.999 118 603 029 311 389 696 × 2 = 1 + 0.998 237 206 058 622 779 392;
  • 54) 0.998 237 206 058 622 779 392 × 2 = 1 + 0.996 474 412 117 245 558 784;

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:


0.000 000 000 742 147 676 451(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 451(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:


0.000 000 000 742 147 676 451(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 -0.000 000 000 742 147 676 451 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