-0.000 000 000 742 147 676 591 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 591 1(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
-0.000 000 000 742 147 676 591 1(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:

|-0.000 000 000 742 147 676 591 1| = 0.000 000 000 742 147 676 591 1


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: 0.000 000 000 742 147 676 591 1.

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) 0.000 000 000 742 147 676 591 1 × 2 = 0 + 0.000 000 001 484 295 353 182 2;
  • 2) 0.000 000 001 484 295 353 182 2 × 2 = 0 + 0.000 000 002 968 590 706 364 4;
  • 3) 0.000 000 002 968 590 706 364 4 × 2 = 0 + 0.000 000 005 937 181 412 728 8;
  • 4) 0.000 000 005 937 181 412 728 8 × 2 = 0 + 0.000 000 011 874 362 825 457 6;
  • 5) 0.000 000 011 874 362 825 457 6 × 2 = 0 + 0.000 000 023 748 725 650 915 2;
  • 6) 0.000 000 023 748 725 650 915 2 × 2 = 0 + 0.000 000 047 497 451 301 830 4;
  • 7) 0.000 000 047 497 451 301 830 4 × 2 = 0 + 0.000 000 094 994 902 603 660 8;
  • 8) 0.000 000 094 994 902 603 660 8 × 2 = 0 + 0.000 000 189 989 805 207 321 6;
  • 9) 0.000 000 189 989 805 207 321 6 × 2 = 0 + 0.000 000 379 979 610 414 643 2;
  • 10) 0.000 000 379 979 610 414 643 2 × 2 = 0 + 0.000 000 759 959 220 829 286 4;
  • 11) 0.000 000 759 959 220 829 286 4 × 2 = 0 + 0.000 001 519 918 441 658 572 8;
  • 12) 0.000 001 519 918 441 658 572 8 × 2 = 0 + 0.000 003 039 836 883 317 145 6;
  • 13) 0.000 003 039 836 883 317 145 6 × 2 = 0 + 0.000 006 079 673 766 634 291 2;
  • 14) 0.000 006 079 673 766 634 291 2 × 2 = 0 + 0.000 012 159 347 533 268 582 4;
  • 15) 0.000 012 159 347 533 268 582 4 × 2 = 0 + 0.000 024 318 695 066 537 164 8;
  • 16) 0.000 024 318 695 066 537 164 8 × 2 = 0 + 0.000 048 637 390 133 074 329 6;
  • 17) 0.000 048 637 390 133 074 329 6 × 2 = 0 + 0.000 097 274 780 266 148 659 2;
  • 18) 0.000 097 274 780 266 148 659 2 × 2 = 0 + 0.000 194 549 560 532 297 318 4;
  • 19) 0.000 194 549 560 532 297 318 4 × 2 = 0 + 0.000 389 099 121 064 594 636 8;
  • 20) 0.000 389 099 121 064 594 636 8 × 2 = 0 + 0.000 778 198 242 129 189 273 6;
  • 21) 0.000 778 198 242 129 189 273 6 × 2 = 0 + 0.001 556 396 484 258 378 547 2;
  • 22) 0.001 556 396 484 258 378 547 2 × 2 = 0 + 0.003 112 792 968 516 757 094 4;
  • 23) 0.003 112 792 968 516 757 094 4 × 2 = 0 + 0.006 225 585 937 033 514 188 8;
  • 24) 0.006 225 585 937 033 514 188 8 × 2 = 0 + 0.012 451 171 874 067 028 377 6;
  • 25) 0.012 451 171 874 067 028 377 6 × 2 = 0 + 0.024 902 343 748 134 056 755 2;
  • 26) 0.024 902 343 748 134 056 755 2 × 2 = 0 + 0.049 804 687 496 268 113 510 4;
  • 27) 0.049 804 687 496 268 113 510 4 × 2 = 0 + 0.099 609 374 992 536 227 020 8;
  • 28) 0.099 609 374 992 536 227 020 8 × 2 = 0 + 0.199 218 749 985 072 454 041 6;
  • 29) 0.199 218 749 985 072 454 041 6 × 2 = 0 + 0.398 437 499 970 144 908 083 2;
  • 30) 0.398 437 499 970 144 908 083 2 × 2 = 0 + 0.796 874 999 940 289 816 166 4;
  • 31) 0.796 874 999 940 289 816 166 4 × 2 = 1 + 0.593 749 999 880 579 632 332 8;
  • 32) 0.593 749 999 880 579 632 332 8 × 2 = 1 + 0.187 499 999 761 159 264 665 6;
  • 33) 0.187 499 999 761 159 264 665 6 × 2 = 0 + 0.374 999 999 522 318 529 331 2;
  • 34) 0.374 999 999 522 318 529 331 2 × 2 = 0 + 0.749 999 999 044 637 058 662 4;
  • 35) 0.749 999 999 044 637 058 662 4 × 2 = 1 + 0.499 999 998 089 274 117 324 8;
  • 36) 0.499 999 998 089 274 117 324 8 × 2 = 0 + 0.999 999 996 178 548 234 649 6;
  • 37) 0.999 999 996 178 548 234 649 6 × 2 = 1 + 0.999 999 992 357 096 469 299 2;
  • 38) 0.999 999 992 357 096 469 299 2 × 2 = 1 + 0.999 999 984 714 192 938 598 4;
  • 39) 0.999 999 984 714 192 938 598 4 × 2 = 1 + 0.999 999 969 428 385 877 196 8;
  • 40) 0.999 999 969 428 385 877 196 8 × 2 = 1 + 0.999 999 938 856 771 754 393 6;
  • 41) 0.999 999 938 856 771 754 393 6 × 2 = 1 + 0.999 999 877 713 543 508 787 2;
  • 42) 0.999 999 877 713 543 508 787 2 × 2 = 1 + 0.999 999 755 427 087 017 574 4;
  • 43) 0.999 999 755 427 087 017 574 4 × 2 = 1 + 0.999 999 510 854 174 035 148 8;
  • 44) 0.999 999 510 854 174 035 148 8 × 2 = 1 + 0.999 999 021 708 348 070 297 6;
  • 45) 0.999 999 021 708 348 070 297 6 × 2 = 1 + 0.999 998 043 416 696 140 595 2;
  • 46) 0.999 998 043 416 696 140 595 2 × 2 = 1 + 0.999 996 086 833 392 281 190 4;
  • 47) 0.999 996 086 833 392 281 190 4 × 2 = 1 + 0.999 992 173 666 784 562 380 8;
  • 48) 0.999 992 173 666 784 562 380 8 × 2 = 1 + 0.999 984 347 333 569 124 761 6;
  • 49) 0.999 984 347 333 569 124 761 6 × 2 = 1 + 0.999 968 694 667 138 249 523 2;
  • 50) 0.999 968 694 667 138 249 523 2 × 2 = 1 + 0.999 937 389 334 276 499 046 4;
  • 51) 0.999 937 389 334 276 499 046 4 × 2 = 1 + 0.999 874 778 668 552 998 092 8;
  • 52) 0.999 874 778 668 552 998 092 8 × 2 = 1 + 0.999 749 557 337 105 996 185 6;
  • 53) 0.999 749 557 337 105 996 185 6 × 2 = 1 + 0.999 499 114 674 211 992 371 2;
  • 54) 0.999 499 114 674 211 992 371 2 × 2 = 1 + 0.998 998 229 348 423 984 742 4;

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:


0.000 000 000 742 147 676 591 1(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 591 1(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:


0.000 000 000 742 147 676 591 1(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 -0.000 000 000 742 147 676 591 1 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