-0.000 000 000 742 147 676 610 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 610 5(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
-0.000 000 000 742 147 676 610 5(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:

|-0.000 000 000 742 147 676 610 5| = 0.000 000 000 742 147 676 610 5


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: 0.000 000 000 742 147 676 610 5.

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) 0.000 000 000 742 147 676 610 5 × 2 = 0 + 0.000 000 001 484 295 353 221;
  • 2) 0.000 000 001 484 295 353 221 × 2 = 0 + 0.000 000 002 968 590 706 442;
  • 3) 0.000 000 002 968 590 706 442 × 2 = 0 + 0.000 000 005 937 181 412 884;
  • 4) 0.000 000 005 937 181 412 884 × 2 = 0 + 0.000 000 011 874 362 825 768;
  • 5) 0.000 000 011 874 362 825 768 × 2 = 0 + 0.000 000 023 748 725 651 536;
  • 6) 0.000 000 023 748 725 651 536 × 2 = 0 + 0.000 000 047 497 451 303 072;
  • 7) 0.000 000 047 497 451 303 072 × 2 = 0 + 0.000 000 094 994 902 606 144;
  • 8) 0.000 000 094 994 902 606 144 × 2 = 0 + 0.000 000 189 989 805 212 288;
  • 9) 0.000 000 189 989 805 212 288 × 2 = 0 + 0.000 000 379 979 610 424 576;
  • 10) 0.000 000 379 979 610 424 576 × 2 = 0 + 0.000 000 759 959 220 849 152;
  • 11) 0.000 000 759 959 220 849 152 × 2 = 0 + 0.000 001 519 918 441 698 304;
  • 12) 0.000 001 519 918 441 698 304 × 2 = 0 + 0.000 003 039 836 883 396 608;
  • 13) 0.000 003 039 836 883 396 608 × 2 = 0 + 0.000 006 079 673 766 793 216;
  • 14) 0.000 006 079 673 766 793 216 × 2 = 0 + 0.000 012 159 347 533 586 432;
  • 15) 0.000 012 159 347 533 586 432 × 2 = 0 + 0.000 024 318 695 067 172 864;
  • 16) 0.000 024 318 695 067 172 864 × 2 = 0 + 0.000 048 637 390 134 345 728;
  • 17) 0.000 048 637 390 134 345 728 × 2 = 0 + 0.000 097 274 780 268 691 456;
  • 18) 0.000 097 274 780 268 691 456 × 2 = 0 + 0.000 194 549 560 537 382 912;
  • 19) 0.000 194 549 560 537 382 912 × 2 = 0 + 0.000 389 099 121 074 765 824;
  • 20) 0.000 389 099 121 074 765 824 × 2 = 0 + 0.000 778 198 242 149 531 648;
  • 21) 0.000 778 198 242 149 531 648 × 2 = 0 + 0.001 556 396 484 299 063 296;
  • 22) 0.001 556 396 484 299 063 296 × 2 = 0 + 0.003 112 792 968 598 126 592;
  • 23) 0.003 112 792 968 598 126 592 × 2 = 0 + 0.006 225 585 937 196 253 184;
  • 24) 0.006 225 585 937 196 253 184 × 2 = 0 + 0.012 451 171 874 392 506 368;
  • 25) 0.012 451 171 874 392 506 368 × 2 = 0 + 0.024 902 343 748 785 012 736;
  • 26) 0.024 902 343 748 785 012 736 × 2 = 0 + 0.049 804 687 497 570 025 472;
  • 27) 0.049 804 687 497 570 025 472 × 2 = 0 + 0.099 609 374 995 140 050 944;
  • 28) 0.099 609 374 995 140 050 944 × 2 = 0 + 0.199 218 749 990 280 101 888;
  • 29) 0.199 218 749 990 280 101 888 × 2 = 0 + 0.398 437 499 980 560 203 776;
  • 30) 0.398 437 499 980 560 203 776 × 2 = 0 + 0.796 874 999 961 120 407 552;
  • 31) 0.796 874 999 961 120 407 552 × 2 = 1 + 0.593 749 999 922 240 815 104;
  • 32) 0.593 749 999 922 240 815 104 × 2 = 1 + 0.187 499 999 844 481 630 208;
  • 33) 0.187 499 999 844 481 630 208 × 2 = 0 + 0.374 999 999 688 963 260 416;
  • 34) 0.374 999 999 688 963 260 416 × 2 = 0 + 0.749 999 999 377 926 520 832;
  • 35) 0.749 999 999 377 926 520 832 × 2 = 1 + 0.499 999 998 755 853 041 664;
  • 36) 0.499 999 998 755 853 041 664 × 2 = 0 + 0.999 999 997 511 706 083 328;
  • 37) 0.999 999 997 511 706 083 328 × 2 = 1 + 0.999 999 995 023 412 166 656;
  • 38) 0.999 999 995 023 412 166 656 × 2 = 1 + 0.999 999 990 046 824 333 312;
  • 39) 0.999 999 990 046 824 333 312 × 2 = 1 + 0.999 999 980 093 648 666 624;
  • 40) 0.999 999 980 093 648 666 624 × 2 = 1 + 0.999 999 960 187 297 333 248;
  • 41) 0.999 999 960 187 297 333 248 × 2 = 1 + 0.999 999 920 374 594 666 496;
  • 42) 0.999 999 920 374 594 666 496 × 2 = 1 + 0.999 999 840 749 189 332 992;
  • 43) 0.999 999 840 749 189 332 992 × 2 = 1 + 0.999 999 681 498 378 665 984;
  • 44) 0.999 999 681 498 378 665 984 × 2 = 1 + 0.999 999 362 996 757 331 968;
  • 45) 0.999 999 362 996 757 331 968 × 2 = 1 + 0.999 998 725 993 514 663 936;
  • 46) 0.999 998 725 993 514 663 936 × 2 = 1 + 0.999 997 451 987 029 327 872;
  • 47) 0.999 997 451 987 029 327 872 × 2 = 1 + 0.999 994 903 974 058 655 744;
  • 48) 0.999 994 903 974 058 655 744 × 2 = 1 + 0.999 989 807 948 117 311 488;
  • 49) 0.999 989 807 948 117 311 488 × 2 = 1 + 0.999 979 615 896 234 622 976;
  • 50) 0.999 979 615 896 234 622 976 × 2 = 1 + 0.999 959 231 792 469 245 952;
  • 51) 0.999 959 231 792 469 245 952 × 2 = 1 + 0.999 918 463 584 938 491 904;
  • 52) 0.999 918 463 584 938 491 904 × 2 = 1 + 0.999 836 927 169 876 983 808;
  • 53) 0.999 836 927 169 876 983 808 × 2 = 1 + 0.999 673 854 339 753 967 616;
  • 54) 0.999 673 854 339 753 967 616 × 2 = 1 + 0.999 347 708 679 507 935 232;

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:


0.000 000 000 742 147 676 610 5(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 610 5(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:


0.000 000 000 742 147 676 610 5(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 -0.000 000 000 742 147 676 610 5 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