-0.000 000 000 742 147 676 459 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 459(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
-0.000 000 000 742 147 676 459(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:

|-0.000 000 000 742 147 676 459| = 0.000 000 000 742 147 676 459


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: 0.000 000 000 742 147 676 459.

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) 0.000 000 000 742 147 676 459 × 2 = 0 + 0.000 000 001 484 295 352 918;
  • 2) 0.000 000 001 484 295 352 918 × 2 = 0 + 0.000 000 002 968 590 705 836;
  • 3) 0.000 000 002 968 590 705 836 × 2 = 0 + 0.000 000 005 937 181 411 672;
  • 4) 0.000 000 005 937 181 411 672 × 2 = 0 + 0.000 000 011 874 362 823 344;
  • 5) 0.000 000 011 874 362 823 344 × 2 = 0 + 0.000 000 023 748 725 646 688;
  • 6) 0.000 000 023 748 725 646 688 × 2 = 0 + 0.000 000 047 497 451 293 376;
  • 7) 0.000 000 047 497 451 293 376 × 2 = 0 + 0.000 000 094 994 902 586 752;
  • 8) 0.000 000 094 994 902 586 752 × 2 = 0 + 0.000 000 189 989 805 173 504;
  • 9) 0.000 000 189 989 805 173 504 × 2 = 0 + 0.000 000 379 979 610 347 008;
  • 10) 0.000 000 379 979 610 347 008 × 2 = 0 + 0.000 000 759 959 220 694 016;
  • 11) 0.000 000 759 959 220 694 016 × 2 = 0 + 0.000 001 519 918 441 388 032;
  • 12) 0.000 001 519 918 441 388 032 × 2 = 0 + 0.000 003 039 836 882 776 064;
  • 13) 0.000 003 039 836 882 776 064 × 2 = 0 + 0.000 006 079 673 765 552 128;
  • 14) 0.000 006 079 673 765 552 128 × 2 = 0 + 0.000 012 159 347 531 104 256;
  • 15) 0.000 012 159 347 531 104 256 × 2 = 0 + 0.000 024 318 695 062 208 512;
  • 16) 0.000 024 318 695 062 208 512 × 2 = 0 + 0.000 048 637 390 124 417 024;
  • 17) 0.000 048 637 390 124 417 024 × 2 = 0 + 0.000 097 274 780 248 834 048;
  • 18) 0.000 097 274 780 248 834 048 × 2 = 0 + 0.000 194 549 560 497 668 096;
  • 19) 0.000 194 549 560 497 668 096 × 2 = 0 + 0.000 389 099 120 995 336 192;
  • 20) 0.000 389 099 120 995 336 192 × 2 = 0 + 0.000 778 198 241 990 672 384;
  • 21) 0.000 778 198 241 990 672 384 × 2 = 0 + 0.001 556 396 483 981 344 768;
  • 22) 0.001 556 396 483 981 344 768 × 2 = 0 + 0.003 112 792 967 962 689 536;
  • 23) 0.003 112 792 967 962 689 536 × 2 = 0 + 0.006 225 585 935 925 379 072;
  • 24) 0.006 225 585 935 925 379 072 × 2 = 0 + 0.012 451 171 871 850 758 144;
  • 25) 0.012 451 171 871 850 758 144 × 2 = 0 + 0.024 902 343 743 701 516 288;
  • 26) 0.024 902 343 743 701 516 288 × 2 = 0 + 0.049 804 687 487 403 032 576;
  • 27) 0.049 804 687 487 403 032 576 × 2 = 0 + 0.099 609 374 974 806 065 152;
  • 28) 0.099 609 374 974 806 065 152 × 2 = 0 + 0.199 218 749 949 612 130 304;
  • 29) 0.199 218 749 949 612 130 304 × 2 = 0 + 0.398 437 499 899 224 260 608;
  • 30) 0.398 437 499 899 224 260 608 × 2 = 0 + 0.796 874 999 798 448 521 216;
  • 31) 0.796 874 999 798 448 521 216 × 2 = 1 + 0.593 749 999 596 897 042 432;
  • 32) 0.593 749 999 596 897 042 432 × 2 = 1 + 0.187 499 999 193 794 084 864;
  • 33) 0.187 499 999 193 794 084 864 × 2 = 0 + 0.374 999 998 387 588 169 728;
  • 34) 0.374 999 998 387 588 169 728 × 2 = 0 + 0.749 999 996 775 176 339 456;
  • 35) 0.749 999 996 775 176 339 456 × 2 = 1 + 0.499 999 993 550 352 678 912;
  • 36) 0.499 999 993 550 352 678 912 × 2 = 0 + 0.999 999 987 100 705 357 824;
  • 37) 0.999 999 987 100 705 357 824 × 2 = 1 + 0.999 999 974 201 410 715 648;
  • 38) 0.999 999 974 201 410 715 648 × 2 = 1 + 0.999 999 948 402 821 431 296;
  • 39) 0.999 999 948 402 821 431 296 × 2 = 1 + 0.999 999 896 805 642 862 592;
  • 40) 0.999 999 896 805 642 862 592 × 2 = 1 + 0.999 999 793 611 285 725 184;
  • 41) 0.999 999 793 611 285 725 184 × 2 = 1 + 0.999 999 587 222 571 450 368;
  • 42) 0.999 999 587 222 571 450 368 × 2 = 1 + 0.999 999 174 445 142 900 736;
  • 43) 0.999 999 174 445 142 900 736 × 2 = 1 + 0.999 998 348 890 285 801 472;
  • 44) 0.999 998 348 890 285 801 472 × 2 = 1 + 0.999 996 697 780 571 602 944;
  • 45) 0.999 996 697 780 571 602 944 × 2 = 1 + 0.999 993 395 561 143 205 888;
  • 46) 0.999 993 395 561 143 205 888 × 2 = 1 + 0.999 986 791 122 286 411 776;
  • 47) 0.999 986 791 122 286 411 776 × 2 = 1 + 0.999 973 582 244 572 823 552;
  • 48) 0.999 973 582 244 572 823 552 × 2 = 1 + 0.999 947 164 489 145 647 104;
  • 49) 0.999 947 164 489 145 647 104 × 2 = 1 + 0.999 894 328 978 291 294 208;
  • 50) 0.999 894 328 978 291 294 208 × 2 = 1 + 0.999 788 657 956 582 588 416;
  • 51) 0.999 788 657 956 582 588 416 × 2 = 1 + 0.999 577 315 913 165 176 832;
  • 52) 0.999 577 315 913 165 176 832 × 2 = 1 + 0.999 154 631 826 330 353 664;
  • 53) 0.999 154 631 826 330 353 664 × 2 = 1 + 0.998 309 263 652 660 707 328;
  • 54) 0.998 309 263 652 660 707 328 × 2 = 1 + 0.996 618 527 305 321 414 656;

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:


0.000 000 000 742 147 676 459(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 459(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:


0.000 000 000 742 147 676 459(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 -0.000 000 000 742 147 676 459 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 54 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