3.141 592 653 589 793 238 462 643 383 279 502 884 197 169 399 375 105 821 007 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 589 793 238 462 643 383 279 502 884 197 169 399 375 105 821 007 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
3.141 592 653 589 793 238 462 643 383 279 502 884 197 169 399 375 105 821 007 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. First, convert to binary (in base 2) the integer part: 3.
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;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

2. 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.

3(10) =

11(2)


3. Convert to binary (base 2) the fractional part: 0.141 592 653 589 793 238 462 643 383 279 502 884 197 169 399 375 105 821 007 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) 0.141 592 653 589 793 238 462 643 383 279 502 884 197 169 399 375 105 821 007 5 × 2 = 0 + 0.283 185 307 179 586 476 925 286 766 559 005 768 394 338 798 750 211 642 015;
  • 2) 0.283 185 307 179 586 476 925 286 766 559 005 768 394 338 798 750 211 642 015 × 2 = 0 + 0.566 370 614 359 172 953 850 573 533 118 011 536 788 677 597 500 423 284 03;
  • 3) 0.566 370 614 359 172 953 850 573 533 118 011 536 788 677 597 500 423 284 03 × 2 = 1 + 0.132 741 228 718 345 907 701 147 066 236 023 073 577 355 195 000 846 568 06;
  • 4) 0.132 741 228 718 345 907 701 147 066 236 023 073 577 355 195 000 846 568 06 × 2 = 0 + 0.265 482 457 436 691 815 402 294 132 472 046 147 154 710 390 001 693 136 12;
  • 5) 0.265 482 457 436 691 815 402 294 132 472 046 147 154 710 390 001 693 136 12 × 2 = 0 + 0.530 964 914 873 383 630 804 588 264 944 092 294 309 420 780 003 386 272 24;
  • 6) 0.530 964 914 873 383 630 804 588 264 944 092 294 309 420 780 003 386 272 24 × 2 = 1 + 0.061 929 829 746 767 261 609 176 529 888 184 588 618 841 560 006 772 544 48;
  • 7) 0.061 929 829 746 767 261 609 176 529 888 184 588 618 841 560 006 772 544 48 × 2 = 0 + 0.123 859 659 493 534 523 218 353 059 776 369 177 237 683 120 013 545 088 96;
  • 8) 0.123 859 659 493 534 523 218 353 059 776 369 177 237 683 120 013 545 088 96 × 2 = 0 + 0.247 719 318 987 069 046 436 706 119 552 738 354 475 366 240 027 090 177 92;
  • 9) 0.247 719 318 987 069 046 436 706 119 552 738 354 475 366 240 027 090 177 92 × 2 = 0 + 0.495 438 637 974 138 092 873 412 239 105 476 708 950 732 480 054 180 355 84;
  • 10) 0.495 438 637 974 138 092 873 412 239 105 476 708 950 732 480 054 180 355 84 × 2 = 0 + 0.990 877 275 948 276 185 746 824 478 210 953 417 901 464 960 108 360 711 68;
  • 11) 0.990 877 275 948 276 185 746 824 478 210 953 417 901 464 960 108 360 711 68 × 2 = 1 + 0.981 754 551 896 552 371 493 648 956 421 906 835 802 929 920 216 721 423 36;
  • 12) 0.981 754 551 896 552 371 493 648 956 421 906 835 802 929 920 216 721 423 36 × 2 = 1 + 0.963 509 103 793 104 742 987 297 912 843 813 671 605 859 840 433 442 846 72;
  • 13) 0.963 509 103 793 104 742 987 297 912 843 813 671 605 859 840 433 442 846 72 × 2 = 1 + 0.927 018 207 586 209 485 974 595 825 687 627 343 211 719 680 866 885 693 44;
  • 14) 0.927 018 207 586 209 485 974 595 825 687 627 343 211 719 680 866 885 693 44 × 2 = 1 + 0.854 036 415 172 418 971 949 191 651 375 254 686 423 439 361 733 771 386 88;
  • 15) 0.854 036 415 172 418 971 949 191 651 375 254 686 423 439 361 733 771 386 88 × 2 = 1 + 0.708 072 830 344 837 943 898 383 302 750 509 372 846 878 723 467 542 773 76;
  • 16) 0.708 072 830 344 837 943 898 383 302 750 509 372 846 878 723 467 542 773 76 × 2 = 1 + 0.416 145 660 689 675 887 796 766 605 501 018 745 693 757 446 935 085 547 52;
  • 17) 0.416 145 660 689 675 887 796 766 605 501 018 745 693 757 446 935 085 547 52 × 2 = 0 + 0.832 291 321 379 351 775 593 533 211 002 037 491 387 514 893 870 171 095 04;
  • 18) 0.832 291 321 379 351 775 593 533 211 002 037 491 387 514 893 870 171 095 04 × 2 = 1 + 0.664 582 642 758 703 551 187 066 422 004 074 982 775 029 787 740 342 190 08;
  • 19) 0.664 582 642 758 703 551 187 066 422 004 074 982 775 029 787 740 342 190 08 × 2 = 1 + 0.329 165 285 517 407 102 374 132 844 008 149 965 550 059 575 480 684 380 16;
  • 20) 0.329 165 285 517 407 102 374 132 844 008 149 965 550 059 575 480 684 380 16 × 2 = 0 + 0.658 330 571 034 814 204 748 265 688 016 299 931 100 119 150 961 368 760 32;
  • 21) 0.658 330 571 034 814 204 748 265 688 016 299 931 100 119 150 961 368 760 32 × 2 = 1 + 0.316 661 142 069 628 409 496 531 376 032 599 862 200 238 301 922 737 520 64;
  • 22) 0.316 661 142 069 628 409 496 531 376 032 599 862 200 238 301 922 737 520 64 × 2 = 0 + 0.633 322 284 139 256 818 993 062 752 065 199 724 400 476 603 845 475 041 28;
  • 23) 0.633 322 284 139 256 818 993 062 752 065 199 724 400 476 603 845 475 041 28 × 2 = 1 + 0.266 644 568 278 513 637 986 125 504 130 399 448 800 953 207 690 950 082 56;
  • 24) 0.266 644 568 278 513 637 986 125 504 130 399 448 800 953 207 690 950 082 56 × 2 = 0 + 0.533 289 136 557 027 275 972 251 008 260 798 897 601 906 415 381 900 165 12;

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).


4. 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:


0.141 592 653 589 793 238 462 643 383 279 502 884 197 169 399 375 105 821 007 5(10) =

0.0010 0100 0011 1111 0110 1010(2)

5. Positive number before normalization:

3.141 592 653 589 793 238 462 643 383 279 502 884 197 169 399 375 105 821 007 5(10) =

11.0010 0100 0011 1111 0110 1010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


3.141 592 653 589 793 238 462 643 383 279 502 884 197 169 399 375 105 821 007 5(10) =

11.0010 0100 0011 1111 0110 1010(2) =

11.0010 0100 0011 1111 0110 1010(2) × 20 =

1.1001 0010 0001 1111 1011 0101 0(2) × 21


7. 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 0 (a positive number)

Exponent (unadjusted): 1

Mantissa (not normalized):
1.1001 0010 0001 1111 1011 0101 0


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

1 + 2(8-1) - 1 =

(1 + 127)(10) =

128(10)


9. 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;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

128(10) =

1000 0000(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 100 1001 0000 1111 1101 1010 10 =

100 1001 0000 1111 1101 1010


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
1000 0000

Mantissa (23 bits) =
100 1001 0000 1111 1101 1010


Decimal number 3.141 592 653 589 793 238 462 643 383 279 502 884 197 169 399 375 105 821 007 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 0000 - 100 1001 0000 1111 1101 1010

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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